Transcript g - Academic Computer Center
Slide 1
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
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Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
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Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
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Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
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Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
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Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
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Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
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Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Copyright 2011 Pearson Education, Inc.
Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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Copyright 2011 Pearson Education, Inc.
The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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Copyright 2011 Pearson Education, Inc.
The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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Copyright 2011 Pearson Education, Inc.
The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Copyright 2011 Pearson Education, Inc.
Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 2
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Copyright 2011 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
Tro: Chemistry: A Molecular Approach, 2/e
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A Small Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
between Kp and Kc
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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Copyright 2011 Pearson Education, Inc.
The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 3
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
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Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Copyright 2011 Pearson Education, Inc.
Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
Tro: Chemistry: A Molecular Approach, 2/e
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
Tro: Chemistry: A Molecular Approach, 2/e
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
Tro: Chemistry: A Molecular Approach, 2/e
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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Copyright 2011 Pearson Education, Inc.
The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
Tro: Chemistry: A Molecular Approach, 2/e
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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Copyright 2011 Pearson Education, Inc.
The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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Copyright 2011 Pearson Education, Inc.
The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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Copyright 2011 Pearson Education, Inc.
The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Copyright 2011 Pearson Education, Inc.
Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 4
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Copyright 2011 Pearson Education, Inc.
Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 5
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Copyright 2011 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
between Kp and Kc
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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33
0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Copyright 2011 Pearson Education, Inc.
Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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90
Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 6
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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Copyright 2011 Pearson Education, Inc.
The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Copyright 2011 Pearson Education, Inc.
Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 7
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
Tro: Chemistry: A Molecular Approach, 2/e
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 8
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
Tro: Chemistry: A Molecular Approach, 2/e
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43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
Tro: Chemistry: A Molecular Approach, 2/e
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
Tro: Chemistry: A Molecular Approach, 2/e
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
Tro: Chemistry: A Molecular Approach, 2/e
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 9
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Copyright 2011 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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Copyright 2011 Pearson Education, Inc.
A Large Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
A Small Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
18
For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Copyright 2011 Pearson Education, Inc.
Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
Tro: Chemistry: A Molecular Approach, 2/e
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Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Copyright 2011 Pearson Education, Inc.
Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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59
+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
Tro: Chemistry: A Molecular Approach, 2/e
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Copyright 2011 Pearson Education, Inc.
Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 10
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 11
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Copyright 2011 Pearson Education, Inc.
Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
Tro: Chemistry: A Molecular Approach, 2/e
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
Tro: Chemistry: A Molecular Approach, 2/e
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Copyright 2011 Pearson Education, Inc.
Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 12
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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A Large Equilibrium Constant
Tro: Chemistry: A Molecular Approach, 2/e
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A Small Equilibrium Constant
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 13
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Copyright 2011 Pearson Education, Inc.
Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
Tro: Chemistry: A Molecular Approach, 2/e
Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
Tro: Chemistry: A Molecular Approach, 2/e
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 14
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
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Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
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Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
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Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
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Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
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Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
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Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
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Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 15
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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43
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 16
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 17
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Copyright 2011 Pearson Education, Inc.
Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 18
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Copyright 2011 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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33
0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Copyright 2011 Pearson Education, Inc.
Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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90
Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 19
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 20
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 21
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
Tro: Chemistry: A Molecular Approach, 2/e
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 22
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Copyright 2011 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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Copyright 2011 Pearson Education, Inc.
A Large Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
A Small Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
18
For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Copyright 2011 Pearson Education, Inc.
Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
Tro: Chemistry: A Molecular Approach, 2/e
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Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Copyright 2011 Pearson Education, Inc.
Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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59
+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
Tro: Chemistry: A Molecular Approach, 2/e
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 23
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 24
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Copyright 2011 Pearson Education, Inc.
Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
Tro: Chemistry: A Molecular Approach, 2/e
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
Tro: Chemistry: A Molecular Approach, 2/e
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 25
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
Tro: Chemistry: A Molecular Approach, 2/e
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
Tro: Chemistry: A Molecular Approach, 2/e
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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A Large Equilibrium Constant
Tro: Chemistry: A Molecular Approach, 2/e
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A Small Equilibrium Constant
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 26
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Copyright 2011 Pearson Education, Inc.
Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
Tro: Chemistry: A Molecular Approach, 2/e
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 27
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
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Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
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Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
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Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
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Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
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Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
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Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
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Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 28
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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43
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 29
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
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Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 30
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 31
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Copyright 2011 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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33
0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Copyright 2011 Pearson Education, Inc.
Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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90
Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 32
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 33
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 34
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
Tro: Chemistry: A Molecular Approach, 2/e
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 35
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Copyright 2011 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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Copyright 2011 Pearson Education, Inc.
A Large Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
A Small Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
18
For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Copyright 2011 Pearson Education, Inc.
Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
Tro: Chemistry: A Molecular Approach, 2/e
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Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Copyright 2011 Pearson Education, Inc.
Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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59
+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
Tro: Chemistry: A Molecular Approach, 2/e
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 36
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 37
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Copyright 2011 Pearson Education, Inc.
Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
Tro: Chemistry: A Molecular Approach, 2/e
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
Tro: Chemistry: A Molecular Approach, 2/e
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
Tro: Chemistry: A Molecular Approach, 2/e
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 38
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
Tro: Chemistry: A Molecular Approach, 2/e
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
Tro: Chemistry: A Molecular Approach, 2/e
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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A Large Equilibrium Constant
Tro: Chemistry: A Molecular Approach, 2/e
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A Small Equilibrium Constant
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 39
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Copyright 2011 Pearson Education, Inc.
Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
Tro: Chemistry: A Molecular Approach, 2/e
Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
Tro: Chemistry: A Molecular Approach, 2/e
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 40
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
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Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
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Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
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Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
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Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
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Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
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Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
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Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 41
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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43
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 42
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
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Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 43
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 44
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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33
0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Copyright 2011 Pearson Education, Inc.
Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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90
Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 45
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 46
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
Tro: Chemistry: A Molecular Approach, 2/e
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 47
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
Tro: Chemistry: A Molecular Approach, 2/e
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43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
Tro: Chemistry: A Molecular Approach, 2/e
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
Tro: Chemistry: A Molecular Approach, 2/e
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
Tro: Chemistry: A Molecular Approach, 2/e
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 48
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Copyright 2011 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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Copyright 2011 Pearson Education, Inc.
A Large Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
A Small Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
18
For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Copyright 2011 Pearson Education, Inc.
Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
Tro: Chemistry: A Molecular Approach, 2/e
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Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Copyright 2011 Pearson Education, Inc.
Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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59
+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
Tro: Chemistry: A Molecular Approach, 2/e
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 49
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 50
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Copyright 2011 Pearson Education, Inc.
Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
Tro: Chemistry: A Molecular Approach, 2/e
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
Tro: Chemistry: A Molecular Approach, 2/e
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
Tro: Chemistry: A Molecular Approach, 2/e
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 51
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
Tro: Chemistry: A Molecular Approach, 2/e
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
Tro: Chemistry: A Molecular Approach, 2/e
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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A Large Equilibrium Constant
Tro: Chemistry: A Molecular Approach, 2/e
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A Small Equilibrium Constant
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 52
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Copyright 2011 Pearson Education, Inc.
Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
Tro: Chemistry: A Molecular Approach, 2/e
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
Tro: Chemistry: A Molecular Approach, 2/e
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 53
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
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Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
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Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
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Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
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Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
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Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
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Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
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Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 54
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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43
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 55
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
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Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 56
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Copyright 2011 Pearson Education, Inc.
Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 57
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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A Large Equilibrium Constant
Tro: Chemistry: A Molecular Approach, 2/e
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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33
0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Copyright 2011 Pearson Education, Inc.
Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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90
Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 58
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 59
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 60
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
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Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
Tro: Chemistry: A Molecular Approach, 2/e
Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
Tro: Chemistry: A Molecular Approach, 2/e
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
Tro: Chemistry: A Molecular Approach, 2/e
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
Tro: Chemistry: A Molecular Approach, 2/e
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 61
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Copyright 2011 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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Copyright 2011 Pearson Education, Inc.
A Large Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
A Small Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
18
For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Copyright 2011 Pearson Education, Inc.
Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
Tro: Chemistry: A Molecular Approach, 2/e
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Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Copyright 2011 Pearson Education, Inc.
Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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59
+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
Tro: Chemistry: A Molecular Approach, 2/e
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 62
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 63
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
Tro: Chemistry: A Molecular Approach, 2/e
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
Tro: Chemistry: A Molecular Approach, 2/e
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
Tro: Chemistry: A Molecular Approach, 2/e
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
Tro: Chemistry: A Molecular Approach, 2/e
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
Tro: Chemistry: A Molecular Approach, 2/e
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Copyright 2011 Pearson Education, Inc.
Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 64
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
Tro: Chemistry: A Molecular Approach, 2/e
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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A Large Equilibrium Constant
Tro: Chemistry: A Molecular Approach, 2/e
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A Small Equilibrium Constant
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 65
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Copyright 2011 Pearson Education, Inc.
Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
Tro: Chemistry: A Molecular Approach, 2/e
Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
Tro: Chemistry: A Molecular Approach, 2/e
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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Copyright 2011 Pearson Education, Inc.
The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 66
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
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Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
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Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
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Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
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Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
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Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
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Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
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Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 67
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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43
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 68
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
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Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
Tro: Chemistry: A Molecular Approach, 2/e
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 69
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Copyright 2011 Pearson Education, Inc.
Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 70
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
between Kp and Kc
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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33
0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Copyright 2011 Pearson Education, Inc.
Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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90
Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 71
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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Copyright 2011 Pearson Education, Inc.
The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Copyright 2011 Pearson Education, Inc.
Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 72
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
Tro: Chemistry: A Molecular Approach, 2/e
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 73
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
Tro: Chemistry: A Molecular Approach, 2/e
Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
Tro: Chemistry: A Molecular Approach, 2/e
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
Tro: Chemistry: A Molecular Approach, 2/e
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
Tro: Chemistry: A Molecular Approach, 2/e
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
Tro: Chemistry: A Molecular Approach, 2/e
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
Tro: Chemistry: A Molecular Approach, 2/e
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 74
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Copyright 2011 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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Copyright 2011 Pearson Education, Inc.
A Large Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
A Small Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
18
For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Copyright 2011 Pearson Education, Inc.
Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
Tro: Chemistry: A Molecular Approach, 2/e
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Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Copyright 2011 Pearson Education, Inc.
Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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59
+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
Tro: Chemistry: A Molecular Approach, 2/e
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Copyright 2011 Pearson Education, Inc.
Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 75
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
between Kp and Kc
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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94
shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 76
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Copyright 2011 Pearson Education, Inc.
Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
Tro: Chemistry: A Molecular Approach, 2/e
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
Tro: Chemistry: A Molecular Approach, 2/e
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
Tro: Chemistry: A Molecular Approach, 2/e
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
Tro: Chemistry: A Molecular Approach, 2/e
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
Tro: Chemistry: A Molecular Approach, 2/e
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 77
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
Tro: Chemistry: A Molecular Approach, 2/e
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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A Large Equilibrium Constant
Tro: Chemistry: A Molecular Approach, 2/e
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A Small Equilibrium Constant
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 78
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Copyright 2011 Pearson Education, Inc.
Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
Tro: Chemistry: A Molecular Approach, 2/e
Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
Tro: Chemistry: A Molecular Approach, 2/e
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
Tro: Chemistry: A Molecular Approach, 2/e
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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Copyright 2011 Pearson Education, Inc.
The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 79
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
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Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
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Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
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Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
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Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
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Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
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Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
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Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Copyright 2011 Pearson Education, Inc.
Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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Copyright 2011 Pearson Education, Inc.
The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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Copyright 2011 Pearson Education, Inc.
The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 80
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
between Kp and Kc
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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43
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 81
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
Tro: Chemistry: A Molecular Approach, 2/e
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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Copyright 2011 Pearson Education, Inc.
The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 82
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Copyright 2011 Pearson Education, Inc.
Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Copyright 2011 Pearson Education, Inc.
Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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Copyright 2011 Pearson Education, Inc.
The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 83
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
between Kp and Kc
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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33
0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Copyright 2011 Pearson Education, Inc.
Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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90
Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 84
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Copyright 2011 Pearson Education, Inc.
Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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Copyright 2011 Pearson Education, Inc.
The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Copyright 2011 Pearson Education, Inc.
Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 85
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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A Large Equilibrium Constant
Tro: Chemistry: A Molecular Approach, 2/e
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A Small Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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Copyright 2011 Pearson Education, Inc.
The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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Copyright 2011 Pearson Education, Inc.
The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 86
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
Tro: Chemistry: A Molecular Approach, 2/e
Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
Tro: Chemistry: A Molecular Approach, 2/e
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
Tro: Chemistry: A Molecular Approach, 2/e
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
Tro: Chemistry: A Molecular Approach, 2/e
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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Copyright 2011 Pearson Education, Inc.
The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
Tro: Chemistry: A Molecular Approach, 2/e
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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Copyright 2011 Pearson Education, Inc.
The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 87
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Copyright 2011 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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Copyright 2011 Pearson Education, Inc.
A Large Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
A Small Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
18
For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Copyright 2011 Pearson Education, Inc.
Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
Tro: Chemistry: A Molecular Approach, 2/e
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Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Copyright 2011 Pearson Education, Inc.
Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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59
+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
Tro: Chemistry: A Molecular Approach, 2/e
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Copyright 2011 Pearson Education, Inc.
Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Copyright 2011 Pearson Education, Inc.
Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 88
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
between Kp and Kc
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
Tro: Chemistry: A Molecular Approach, 2/e
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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90
Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
94
shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 89
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
Tro: Chemistry: A Molecular Approach, 2/e
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
Tro: Chemistry: A Molecular Approach, 2/e
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
Tro: Chemistry: A Molecular Approach, 2/e
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
Tro: Chemistry: A Molecular Approach, 2/e
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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Copyright 2011 Pearson Education, Inc.
The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
Tro: Chemistry: A Molecular Approach, 2/e
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 90
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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A Large Equilibrium Constant
Tro: Chemistry: A Molecular Approach, 2/e
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A Small Equilibrium Constant
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 91
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Copyright 2011 Pearson Education, Inc.
Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
Tro: Chemistry: A Molecular Approach, 2/e
Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
Tro: Chemistry: A Molecular Approach, 2/e
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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Copyright 2011 Pearson Education, Inc.
The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 92
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
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Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
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Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
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Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
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Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
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Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
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Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
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Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Copyright 2011 Pearson Education, Inc.
Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 93
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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43
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
94
shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 94
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Copyright 2011 Pearson Education, Inc.
Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
Tro: Chemistry: A Molecular Approach, 2/e
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
Tro: Chemistry: A Molecular Approach, 2/e
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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Copyright 2011 Pearson Education, Inc.
The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
Tro: Chemistry: A Molecular Approach, 2/e
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Copyright 2011 Pearson Education, Inc.
Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 2
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Copyright 2011 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
between Kp and Kc
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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33
0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Copyright 2011 Pearson Education, Inc.
Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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90
Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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94
shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 3
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
Tro: Chemistry: A Molecular Approach, 2/e
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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Copyright 2011 Pearson Education, Inc.
The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 4
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 5
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
Tro: Chemistry: A Molecular Approach, 2/e
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43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
Tro: Chemistry: A Molecular Approach, 2/e
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
Tro: Chemistry: A Molecular Approach, 2/e
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 6
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Copyright 2011 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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Copyright 2011 Pearson Education, Inc.
A Large Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
A Small Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
18
For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Copyright 2011 Pearson Education, Inc.
Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
Tro: Chemistry: A Molecular Approach, 2/e
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Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Copyright 2011 Pearson Education, Inc.
Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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59
+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
Tro: Chemistry: A Molecular Approach, 2/e
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Copyright 2011 Pearson Education, Inc.
Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 7
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
between Kp and Kc
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 8
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
Tro: Chemistry: A Molecular Approach, 2/e
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
Tro: Chemistry: A Molecular Approach, 2/e
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
Tro: Chemistry: A Molecular Approach, 2/e
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Copyright 2011 Pearson Education, Inc.
Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 9
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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A Large Equilibrium Constant
Tro: Chemistry: A Molecular Approach, 2/e
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A Small Equilibrium Constant
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 10
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Copyright 2011 Pearson Education, Inc.
Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
Tro: Chemistry: A Molecular Approach, 2/e
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 11
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
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Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
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Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
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Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
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Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
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Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
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Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
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Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 12
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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43
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 13
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
Tro: Chemistry: A Molecular Approach, 2/e
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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Copyright 2011 Pearson Education, Inc.
The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Copyright 2011 Pearson Education, Inc.
Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 14
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 15
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Copyright 2011 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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33
0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Copyright 2011 Pearson Education, Inc.
Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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90
Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 16
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 17
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 18
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
Tro: Chemistry: A Molecular Approach, 2/e
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43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
Tro: Chemistry: A Molecular Approach, 2/e
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 19
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Copyright 2011 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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Copyright 2011 Pearson Education, Inc.
A Large Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
A Small Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
18
For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Copyright 2011 Pearson Education, Inc.
Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
Tro: Chemistry: A Molecular Approach, 2/e
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Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Copyright 2011 Pearson Education, Inc.
Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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59
+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
Tro: Chemistry: A Molecular Approach, 2/e
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 20
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 21
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
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Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
Tro: Chemistry: A Molecular Approach, 2/e
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
Tro: Chemistry: A Molecular Approach, 2/e
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 22
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
Tro: Chemistry: A Molecular Approach, 2/e
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
Tro: Chemistry: A Molecular Approach, 2/e
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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A Large Equilibrium Constant
Tro: Chemistry: A Molecular Approach, 2/e
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A Small Equilibrium Constant
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 23
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Copyright 2011 Pearson Education, Inc.
Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
Tro: Chemistry: A Molecular Approach, 2/e
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 24
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
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Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
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Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
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Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
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Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
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Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
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Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
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Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 25
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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43
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 26
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
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Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 27
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Copyright 2011 Pearson Education, Inc.
Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 28
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Copyright 2011 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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33
0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Copyright 2011 Pearson Education, Inc.
Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 29
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 30
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
Tro: Chemistry: A Molecular Approach, 2/e
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 31
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
Tro: Chemistry: A Molecular Approach, 2/e
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 32
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Copyright 2011 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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Copyright 2011 Pearson Education, Inc.
A Large Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
A Small Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
18
For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Copyright 2011 Pearson Education, Inc.
Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
Tro: Chemistry: A Molecular Approach, 2/e
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Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Copyright 2011 Pearson Education, Inc.
Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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59
+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
Tro: Chemistry: A Molecular Approach, 2/e
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 33
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 34
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
Tro: Chemistry: A Molecular Approach, 2/e
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
Tro: Chemistry: A Molecular Approach, 2/e
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
Tro: Chemistry: A Molecular Approach, 2/e
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 35
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
Tro: Chemistry: A Molecular Approach, 2/e
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
Tro: Chemistry: A Molecular Approach, 2/e
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
Tro: Chemistry: A Molecular Approach, 2/e
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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A Large Equilibrium Constant
Tro: Chemistry: A Molecular Approach, 2/e
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A Small Equilibrium Constant
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 36
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Copyright 2011 Pearson Education, Inc.
Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
Tro: Chemistry: A Molecular Approach, 2/e
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 37
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
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Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
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Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
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Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
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Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
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Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
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Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
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Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 38
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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43
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 39
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 40
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
Tro: Chemistry: A Molecular Approach, 2/e
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 41
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Copyright 2011 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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33
0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Copyright 2011 Pearson Education, Inc.
Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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90
Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 42
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 43
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
Tro: Chemistry: A Molecular Approach, 2/e
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 44
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
Tro: Chemistry: A Molecular Approach, 2/e
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43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
Tro: Chemistry: A Molecular Approach, 2/e
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 45
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Copyright 2011 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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Copyright 2011 Pearson Education, Inc.
A Large Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
A Small Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
18
For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Copyright 2011 Pearson Education, Inc.
Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
Tro: Chemistry: A Molecular Approach, 2/e
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Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Copyright 2011 Pearson Education, Inc.
Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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59
+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
Tro: Chemistry: A Molecular Approach, 2/e
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 46
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 47
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
Tro: Chemistry: A Molecular Approach, 2/e
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
Tro: Chemistry: A Molecular Approach, 2/e
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
Tro: Chemistry: A Molecular Approach, 2/e
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 48
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
Tro: Chemistry: A Molecular Approach, 2/e
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
Tro: Chemistry: A Molecular Approach, 2/e
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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A Large Equilibrium Constant
Tro: Chemistry: A Molecular Approach, 2/e
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A Small Equilibrium Constant
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 49
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Copyright 2011 Pearson Education, Inc.
Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
Tro: Chemistry: A Molecular Approach, 2/e
Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
Tro: Chemistry: A Molecular Approach, 2/e
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
Tro: Chemistry: A Molecular Approach, 2/e
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 50
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
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Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
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Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
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Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
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Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
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Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
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Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
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Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 51
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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43
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 52
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
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Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 53
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
Tro: Chemistry: A Molecular Approach, 2/e
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 54
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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33
0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Copyright 2011 Pearson Education, Inc.
Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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90
Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 55
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 56
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
Tro: Chemistry: A Molecular Approach, 2/e
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 57
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
Tro: Chemistry: A Molecular Approach, 2/e
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43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
Tro: Chemistry: A Molecular Approach, 2/e
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
Tro: Chemistry: A Molecular Approach, 2/e
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
Tro: Chemistry: A Molecular Approach, 2/e
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 58
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Copyright 2011 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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Copyright 2011 Pearson Education, Inc.
A Large Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
A Small Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
18
For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Copyright 2011 Pearson Education, Inc.
Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
Tro: Chemistry: A Molecular Approach, 2/e
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Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Copyright 2011 Pearson Education, Inc.
Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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59
+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
Tro: Chemistry: A Molecular Approach, 2/e
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 59
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 60
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Copyright 2011 Pearson Education, Inc.
Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
Tro: Chemistry: A Molecular Approach, 2/e
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
Tro: Chemistry: A Molecular Approach, 2/e
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
Tro: Chemistry: A Molecular Approach, 2/e
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 61
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
Tro: Chemistry: A Molecular Approach, 2/e
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
Tro: Chemistry: A Molecular Approach, 2/e
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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A Large Equilibrium Constant
Tro: Chemistry: A Molecular Approach, 2/e
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A Small Equilibrium Constant
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 62
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Copyright 2011 Pearson Education, Inc.
Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
Tro: Chemistry: A Molecular Approach, 2/e
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
Tro: Chemistry: A Molecular Approach, 2/e
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 63
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
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Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
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Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
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Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
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Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
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Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
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Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
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Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Copyright 2011 Pearson Education, Inc.
Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 64
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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43
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 65
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
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Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
Tro: Chemistry: A Molecular Approach, 2/e
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 66
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Copyright 2011 Pearson Education, Inc.
Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 67
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Copyright 2011 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
Tro: Chemistry: A Molecular Approach, 2/e
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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33
0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Copyright 2011 Pearson Education, Inc.
Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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90
Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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94
shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 68
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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Copyright 2011 Pearson Education, Inc.
The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 69
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
Tro: Chemistry: A Molecular Approach, 2/e
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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A Large Equilibrium Constant
Tro: Chemistry: A Molecular Approach, 2/e
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A Small Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 70
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
Tro: Chemistry: A Molecular Approach, 2/e
Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
Tro: Chemistry: A Molecular Approach, 2/e
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
Tro: Chemistry: A Molecular Approach, 2/e
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
Tro: Chemistry: A Molecular Approach, 2/e
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 71
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Copyright 2011 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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Copyright 2011 Pearson Education, Inc.
A Large Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
A Small Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
18
For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Copyright 2011 Pearson Education, Inc.
Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
Tro: Chemistry: A Molecular Approach, 2/e
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Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Copyright 2011 Pearson Education, Inc.
Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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59
+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
Tro: Chemistry: A Molecular Approach, 2/e
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 72
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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94
shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 73
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
Tro: Chemistry: A Molecular Approach, 2/e
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
Tro: Chemistry: A Molecular Approach, 2/e
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
Tro: Chemistry: A Molecular Approach, 2/e
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
Tro: Chemistry: A Molecular Approach, 2/e
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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Copyright 2011 Pearson Education, Inc.
The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
Tro: Chemistry: A Molecular Approach, 2/e
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Copyright 2011 Pearson Education, Inc.
Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 74
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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A Large Equilibrium Constant
Tro: Chemistry: A Molecular Approach, 2/e
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A Small Equilibrium Constant
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 75
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Copyright 2011 Pearson Education, Inc.
Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
Tro: Chemistry: A Molecular Approach, 2/e
Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
Tro: Chemistry: A Molecular Approach, 2/e
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Copyright 2011 Pearson Education, Inc.
Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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Copyright 2011 Pearson Education, Inc.
The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 76
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
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Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
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Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
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Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
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Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
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Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
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Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
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Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Copyright 2011 Pearson Education, Inc.
Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 77
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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43
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 78
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
Tro: Chemistry: A Molecular Approach, 2/e
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 79
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Copyright 2011 Pearson Education, Inc.
Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 80
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
between Kp and Kc
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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33
0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Copyright 2011 Pearson Education, Inc.
Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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90
Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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94
shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 81
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
Tro: Chemistry: A Molecular Approach, 2/e
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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Copyright 2011 Pearson Education, Inc.
The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Copyright 2011 Pearson Education, Inc.
Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 82
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
Tro: Chemistry: A Molecular Approach, 2/e
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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A Large Equilibrium Constant
Tro: Chemistry: A Molecular Approach, 2/e
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A Small Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Copyright 2011 Pearson Education, Inc.
Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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Copyright 2011 Pearson Education, Inc.
The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 83
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
Tro: Chemistry: A Molecular Approach, 2/e
Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
Tro: Chemistry: A Molecular Approach, 2/e
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
Tro: Chemistry: A Molecular Approach, 2/e
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
Tro: Chemistry: A Molecular Approach, 2/e
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
Tro: Chemistry: A Molecular Approach, 2/e
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 84
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Copyright 2011 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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Copyright 2011 Pearson Education, Inc.
A Large Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
A Small Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
18
For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Copyright 2011 Pearson Education, Inc.
Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
Tro: Chemistry: A Molecular Approach, 2/e
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Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Copyright 2011 Pearson Education, Inc.
Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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59
+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
Tro: Chemistry: A Molecular Approach, 2/e
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Copyright 2011 Pearson Education, Inc.
Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Copyright 2011 Pearson Education, Inc.
Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 85
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
Tro: Chemistry: A Molecular Approach, 2/e
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
between Kp and Kc
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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86
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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90
Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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94
shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 86
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
Tro: Chemistry: A Molecular Approach, 2/e
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
Tro: Chemistry: A Molecular Approach, 2/e
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
Tro: Chemistry: A Molecular Approach, 2/e
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
Tro: Chemistry: A Molecular Approach, 2/e
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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Copyright 2011 Pearson Education, Inc.
The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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Copyright 2011 Pearson Education, Inc.
The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Copyright 2011 Pearson Education, Inc.
Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 87
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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A Large Equilibrium Constant
Tro: Chemistry: A Molecular Approach, 2/e
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A Small Equilibrium Constant
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 88
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Copyright 2011 Pearson Education, Inc.
Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
Tro: Chemistry: A Molecular Approach, 2/e
Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
Tro: Chemistry: A Molecular Approach, 2/e
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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Copyright 2011 Pearson Education, Inc.
The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 89
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
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Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
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Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
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Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
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Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
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Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
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Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
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Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Copyright 2011 Pearson Education, Inc.
Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Copyright 2011 Pearson Education, Inc.
Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 90
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
between Kp and Kc
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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43
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
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The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 91
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
Tro: Chemistry: A Molecular Approach, 2/e
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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Copyright 2011 Pearson Education, Inc.
The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Copyright 2011 Pearson Education, Inc.
Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 92
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
Tro: Chemistry: A Molecular Approach, 2/e
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Hypothetical Reaction
2 Red Blue
Tro: Chemistry: A Molecular Approach, 2/e
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
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For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
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Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
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Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Copyright 2011 Pearson Education, Inc.
Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Copyright 2011 Pearson Education, Inc.
Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 93
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Copyright 2011 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
Tro: Chemistry: A Molecular Approach, 2/e
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Copyright 2011 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
between Kp and Kc
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Copyright 2011 Pearson Education, Inc.
Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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33
0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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Copyright 2011 Pearson Education, Inc.
The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Copyright 2011 Pearson Education, Inc.
Q, K, and the Direction of Reaction
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Copyright 2011 Pearson Education, Inc.
Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Copyright 2011 Pearson Education, Inc.
Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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90
Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
94
shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.
Slide 94
Chapter 14
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright 2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two
•
•
kinds of arrows in reactions to
indicate the degree of
completion of the reactions
A single arrow indicates all the
reactant molecules are
converted to product molecules
at the end
A double arrow indicates the
reaction stops when only some
of the reactant molecules have
been converted into products
in these notes
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Copyright 2011 Pearson Education, Inc.
Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made
forward reaction = reactants products
therefore the reactant concentrations decrease and the
product concentrations increase
as reactant concentration decreases, the forward reaction rate
decreases
• Eventually, the products can react to re-form some of
the reactants
reverse reaction = products reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate
increases
• Processes that proceed in both the forward and reverse
direction are said to be reversible
reactants products
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Hypothetical Reaction
2 Red Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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Hypothetical Reaction
2 Red Blue
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Reaction Dynamics
Rate
Once
Eventually,
Because
equilibrium
thethe
reactant
reaction
is established,
concentration
proceeds
As
Initially,
the
forward
only
the
reaction
forward
proceeds
the
in
decreases,
the
forward
reverse
the
and
direction
forward
reverse
as
reaction
reactions
fast asslows.
itproceed
reaction
products
takes
place
andrate,
uses
Asmakes
proceeds
the products
at
the
in the
same
accumulate,
forward
direction.
soreactants
the
concentrations
At
reverse
this time
reaction
equilibrium
ofspeeds
all materials
isup.
established.
stay constant
Rate Forward
Rate Reverse
Time
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Dynamic Equilibrium
• As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the
same rate
• Dynamic equilibrium is the condition wherein
the rates of the forward and reverse reactions
are equal
• Once the reaction reaches equilibrium, the
concentrations of all the chemicals remain
constant
because the chemicals are being consumed and
made at the same rate
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Copyright 2011 Pearson Education, Inc.
Equilibrium Equal
• The rates of the forward and reverse
•
•
•
reactions are equal at equilibrium
But that does not mean the concentrations of
reactants and products are equal
Some reactions reach equilibrium only after
almost all the reactant molecules are
consumed – we say the position of
equilibrium favors the products
Other reactions reach equilibrium when only
a small percentage of the reactant molecules
are consumed – we say the position of
equilibrium favors the reactants
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Copyright 2011 Pearson Education, Inc.
An Analogy:
Population Changes
However,
When Country
after a time,
A citizens
emigration
feel overcrowded,
will occur in
both
some
directions
will emigrate
at the same
to Country
rate, B
leading to
populations in Country A and Country B that are
constant, though not necessarily equal
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Equilibrium Constant
• Even though the concentrations of reactants and
products are not equal at equilibrium, there is a
relationship between them
• The relationship between the chemical equation and
the concentrations of reactants and products is called
the Law of Mass Action
• For the general equation aA + bB cC + dD, the
Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the
balanced chemical equation
always products over reactants
• K is called the equilibrium constant
unitless
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Writing Equilibrium Constant
Expressions
• For the reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
the equilibrium constant
expression is
• So for the reaction
2 N2O5(g) 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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Copyright 2011 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, we know that when
the reaction reaches equilibrium there will be
many more product molecules present than
reactant molecules
the position of equilibrium favors products
• When the value of Keq << 1, we know that when
the reaction reaches equilibrium there will be
many more reactant molecules present than
product molecules
the position of equilibrium favors reactants
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A Large Equilibrium Constant
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A Small Equilibrium Constant
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Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) 2 SO3(g) K = 8 x 1025
favors products
N2(g) + 2 O2(g) 2 NO2(g)
K = 3 x 10−17
favors reactants
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Relationships between K
and Chemical Equations
• When the reaction is written backward, the
equilibrium constant is inverted
For the reaction aA + bB cC + dD
the equilibrium constant expression
is
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For the reaction cC + dD aA + bB
the equilibrium constant expression
is
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor
For the reaction aA + bB cC
the equilibrium constant
expression is
Tro: Chemistry: A Molecular Approach, 2/e
For the reaction 2aA + 2bB 2cC
the equilibrium constant
expression is
17
Copyright 2011 Pearson Education, Inc.
Relationships between K
and Chemical Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations
For the reactions (1) aA bB
and (2) bB cC the equilibrium
constant expressions are
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For the reaction aA cC
the equilibrium constant
expression is
Copyright 2011 Pearson Education, Inc.
Example 14.2: Compute the equilibrium constant at
25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 108 at 25 °C
Find: K for NH3(g) 0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual
K
K’
Plan:
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g) 2 NH3(g)
K1 = 3.7 x 108
2 NH3(g) N2(g) + 3 H2(g)
NH3(g) 0.5 N2(g) + 1.5 H2(g)
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Practice – When the reaction A(aq) 2 B(aq) reaches
equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
When the reaction 2 B(aq) Z(aq) reaches equilibrium
[B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
Calculate the equilibrium constant for each of these
reactions and the equilibrium constant for the
reaction 3 Z(aq) 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2BA
Z2B
ZA
3Z3A
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K3 = 1/Krxn 1 = 6.25 x 10−5
K4 = 1/Krxn 2 = 2
K3K4 = 1.25 x 10−4
(K3K4)3 = 2.0 x 10−12
20
Copyright 2011 Pearson Education, Inc.
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is
proportional to its partial pressure
• Therefore, the equilibrium constant can be
expressed as the ratio of the partial pressures
of the gases
• For aA(g) + bB(g) cC(g) + dD(g) the
equilibrium constant expressions are
or
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Kc and Kp
• In calculating Kp, the partial pressures are
always in atm
• The values of Kp and Kc are not necessarily
the same
because of the difference in units
Kp = Kc when Dn = 0
• The relationship between them is
Dn is the difference between the number of
moles of reactants and moles of products
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Deriving the Relationship
between Kp and Kc
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Copyright 2011 Pearson Education, Inc.
Deriving the Relationship
Between Kp and Kc
for aA(g) + bB(g) cC(g) + dD(g)
substituting
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Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) 2
NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Given: Kp = 2.2 x 1012
Find: Kc
Conceptual
Plan:
Kp
Kc
Relationships:
Solution:
2 NO(g) + O2(g) 2 NO2(g)
Dn = 2 3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
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Copyright 2011 Pearson Education, Inc.
Practice – Calculate the value of Kp or Kc
for each of the following at 27 °C
2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g)
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Kp = 3 x 10−17
Copyright 2011 Pearson Education, Inc.
Heterogeneous Equilibria
• Pure solids and pure liquids are materials whose
concentration doesn’t change during the course of
a reaction
its amount can change, but the amount of it in solution
doesn’t
because it isn’t in solution
• Because their concentration doesn’t change,
•
solids and liquids are not included in the
equilibrium constant expression
For the reaction aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
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Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore the amount
of C has no effect on
the position of
equilibrium.
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Copyright 2011 Pearson Education, Inc.
Practice – Write the equilibrium constant
expressions, K, and predict the position of
equilibrium for the following
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) CaSO4(s) + 2 HNO3(aq)
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K = 1 x 104
Copyright 2011 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium
•
constant is to measure the amounts of reactants
and products in a mixture at equilibrium
The equilibrium mixture may have different
amounts of reactants and products, but the value
of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of
the initial amounts of reactants and products
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Initial and Equilibrium Concentrations for
H2(g) + I2(g) 2HI(g) @ 445 °C
Initial
Concentration
Eq uilibri um
Con centr ation
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
0.50
0.50
0.0
0.11
0.11
0.78
0.0
0.0
0.50 0.055 0.055 0.39
0.50
0.50 0.50 0.165 0.165 1.17
1.0
0.5
0.0
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Equilibrium
Constant
0.53 0.033 0.934
31
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Calculating Equilibrium
Concentrations
• Stoichiometry can be used to determine the
equilibrium concentrations of all reactants
and products if you know initial
concentrations and one equilibrium
concentration
• Use the change in the concentration of the
material that you know to determine the
change in the other chemicals in the
reaction
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Calculating Equilibrium Concentrations:
An Example
A(aq)
+ B(aq)2
4C
• Suppose you 2
have
a reaction
A(aq)
+ (aq)
B(aq) 4 C(aq)
Product C increased
0.50M,
mol/L
with initial concentrations
[A] =by
1.00
[B] = 1.00 M,
Because
Product
C increased
by 0.50 the
mol/L,
you must have
and [C]
= 0. You
then measure
equilibrium
used
½ that amountofofCReactant
¼
A,
according
to the
concentration
as [C] =B,
0.50
M.
stoichiometry.
-½(0.50) -¼(0.50) +0.50
0.75
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0.88
Copyright 2011 Pearson Education, Inc.
Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
+0.035
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Example 14.6: Find the value of Kc for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700 °C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
−2(0.035) +0.035 +3(0.035)
0.045
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0.105
Copyright 2011 Pearson Education, Inc.
Practice –The following data were collected for the
reaction 2 NO2(g) N2O4(g) at 100 °C. Complete the
table and determine values of Kp and Kc.
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Copyright 2011 Pearson Education, Inc.
Practice – A second experiment was done and the following
data were collected for the reaction 2 NO2(g) N2O4(g) at
100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
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The Reaction Quotient
• If a reaction mixture, containing
•
•
both reactants and products, is for the gas phase reaction
not at equilibrium; how can we aA + bB cC + dD
determine in which direction it the reaction quotient is:
will proceed?
The answer is to compare the
current concentration ratios to the
equilibrium constant
The concentration ratio of the
products (raised to the power of
their coefficients) to the reactants
(raised to the power of their
coefficients) is called the reaction
quotient, Q
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The Reaction Quotient:
Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the
reverse direction
the [products] will decrease and [reactants] will increase
• If Q < K, the reaction will proceed fastest in the
forward direction
the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0,
•
and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞,
and the reaction will proceed in the reverse direction
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Q, K, and the Direction of Reaction
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Example 14.7: For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
Given: for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Conceptual P , P , P
Q
I2
Cl2
ICl
Plan:
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) 2 ICl(g)
Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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Practice – For the reaction CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
Kc = 3.59 at 900 °C. For each of the following measured
concentrations determine whether the reaction is at equilibrium.
If not at equilibrium, in which direction will the reaction proceed to
get to equilibrium?
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Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and
Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 CO2 + CF4
reaction and Kc. You’re
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
also given the [X]eq of all
Find: [CO2]eq
but one of the chemicals
Conceptual
Strategize: You can
[CO2]
Plan: K, [COF2], [CF4]
calculate the missing
concentration by using
the equilibrium constant Relationships:
expression
Solution:
Solve: Solve the
equilibrium constant
expression for the
unknown quantity by
substituting in the given
amounts
Check: substitute
approximations back in
to see if the value agrees
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Check:
43
Copyright 2011 Pearson Education, Inc.
Practice – A sample of PCl5(g) is placed in a 0.500 L
container and heated to 160 °C. The PCl5 is decomposed
into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3
and Cl2 are formed. Determine the equilibrium concentration
of PCl5 if Kc = 0.0635
PCl5 PCl3 + Cl2
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0.203 m ol
0.203 m ol
0.500 L
0.500 L
Copyright 2011 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the
coefficient of x
the x change is + for materials on the side the reaction is
proceeding toward
the x change is for materials on the side the reaction is
proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the
quadratic formula
may be able to simplify and approximate answer for very large or
small equilibrium constants
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Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial
pressures are all 0.100 atm, find the equilibrium
concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
x
x
+2x
0.100x 0.100x 0.100+2x
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729
+2x
x −0.0729
x +2(0.0729)
0.100x
0.027 0.100x
0.027 0.100+2x
0.246
0.0729 x
PI 0.100 x 0.100 0.0729 0.027 atm
2
PC l 0.100 x 0.100 0.0729 0.027 atm
2
PIC l 0.100 2 x 0.100 2 0.0729 0.246 atm
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Copyright 2011 Pearson Education, Inc.
Example 14.11: For the reaction I2(g) + Cl2(g) 2
ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures
are all 0.100 atm, find the equilibrium concentrations
−0.0729 −0.0729 2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
(Hint: you will need to use the quadratic formula to
solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q =
0 and the reaction
must proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00216 = 0.498
[I2] = 0.498 M
2(0.00216) = 0.00432
[I] = 0.00432 M
x = 0.00216
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Approximations to Simplify the Math
• When the equilibrium constant is very small, the
•
position of equilibrium favors the reactants
For relatively large initial concentrations of
reactants, the reactant concentration will not
change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial ax) [X]initial
we are approximating the equilibrium concentration of
reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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Checking the Approximation and
Refining as Necessary
• We can check our approximation by comparing
•
the approximate value of x to the initial
concentration
If the approximate value of x is less than 5% of
the initial concentration, the approximation is
valid
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Example14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find
the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E−4
2.50E−4
2x
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+2x
+x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to 800
°C, find the equilibrium concentrations.
2x
2.50E–4
+2x
2x
+x
x
the approximation is not valid!!
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
xcurrent = 1.38 x
10−5
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2x
+2x
+x
2.50E−4
2x
2x
x
Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
2.50E−4
2x
xcurrent = 1.27 x 10−5
+2x
+x
2x
x
since xcurrent = xnew, approx. OK
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.50E−4
2x
x
2.24E−4 2.56E−5
1.28E−5
2x
xcurrent = 1.28 x 10−5
H 2 S 2.50 10 4 2 x 2.50 10 4 2 1.28 10 5 2.24 10 4 M
H 2 2 x 2 1.28 10 5 2.56 10 5 M
S 2 x 1.28 10 5 M
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Copyright 2011 Pearson Education, Inc.
Example 14.13: For the reaction 2 H2S(g) 2 H2(g) +
S2(g) @ 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask
initially containing 1.25 x 10−4 mol H2S is heated to
800 °C, find the equilibrium concentrations.
2x
+2x
+x
2.24E−4 2.56E−5 1.28E−5
Kc(calculated) = Kc(given) within significant figures
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the
value of Kc = 3.76 x 10−5 at 1000 K. If 1.00
mole of I2 is placed into a 2.00 L flask and
heated, what will be the equilibrium
concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
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Practice – For the reaction I2(g) 2 I(g) the value of
Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0
and the reaction must
proceed forward
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Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction I2(g) 2 I(g) the value
of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is
placed into a 2.00 L flask and heated, what will be
the equilibrium concentrations of [I2] and [I]?
0.500 0.00217 = 0.498
[I2] = 0.498 M
2(0.00217) = 0.00434
[I] = 0.00434 M
x = 0.00217
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial
concentration of N2O4 is 0.0125 M, what will be the
equilibrium concentrations of [N2O4] and [NO2]?
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Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0,
Q = 0 and the reaction
must proceed forward
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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Copyright 2011 Pearson Education, Inc.
Practice – For the reaction N2O4(g) 2 NO2(g) the
value of Kc = 1.07 x 10−5 . If the initial concentration
of N2O4 is 0.0125 M, what will be the equilibrium
concentrations of [N2O4] and [NO2]?
[N2O4]= 0.0125 1.83 x 10−4
[N2O4] = 0.0123 M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4
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Copyright 2011 Pearson Education, Inc.
Disturbing and Restoring
Equilibrium
• Once a reaction is at equilibrium, the
•
•
concentrations of all the reactants and
products remain the same
However if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored
The new concentrations will be different, but
the equilibrium constant will be the same
unless you change the temperature
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Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting
•
the effect various changes in conditions have
on the position of equilibrium
It says that if a system at equilibrium is
disturbed, the position of equilibrium will shift to
minimize the disturbance
disturbances all involve making the system open
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Copyright 2011 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving
from Country B into Country A
faster than people moving from
Country A into Country B. This
will continue until a new
equilibrium between the
When
the populations
Country
A
andisCountry
BB the
populations
established;
When
an influx of of
population
enters
Country
arefrom
in equilibrium,
emigration
rates
the
new
populations
will have
somewherethe
outside
Country
A, itbetween
disturbs
the
two countries
are equal
sodifferent
the populations
stay
constant.
numbers
ofCountry
people
than
equilibrium
established
between
Country
A and
B.
the old ones.
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Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
•
•
forward reaction, but has no initial effect on the
rate of the reverse reaction
The reaction proceeds to the right until equilibrium
is restored
At the new equilibrium position, you will have
more of the products than before, less of the nonadded reactants than before, and less of the
added reactant
but not as little of the added reactant as you had before
the addition
• At the new equilibrium position, the
concentrations of reactants and products will be
such that the value of the equilibrium constant is
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium:
Adding or Removing Reactants
• After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium
constant expression
i.e., not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
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Copyright 2011 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
so the reaction is going faster in reverse
• The reaction proceeds to the left until equilibrium is
•
restored
At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant
but not as much of the removed reactant as you had
before the removal
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same
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Copyright 2011 Pearson Education, Inc.
The Effect of Concentration Changes
on Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added
chemicals or toward side with removed
chemicals
Remember, adding more of a solid or liquid does not
change its concentration – and therefore has no effect on
the equilibrium
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The Effect of Concentration Changes
on Equilibrium
When NO2 is added, some of it
combines to make more N2O4
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Copyright 2011 Pearson Education, Inc.
The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial
pressure, causing the equilibrium to shift to the
right
increasing its partial pressure increases its
concentration
does not increase the partial pressure of the other
gases in the mixture
• Adding an inert gas to the mixture has no effect
on the position of equilibrium
does not affect the partial pressures of the gases in
the reaction
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Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases
the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total
•
•
•
pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure
The way the system reduces the pressure is to
reduce the number of gas molecules in the container
When the volume decreases, the equilibrium
shifts to the side with fewer gas molecules
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Disturbing Equilibrium:
Changing the Volume
• After equilibrium is established, the container
•
•
•
volume is decreased
How will it affect the concentration of solids,
liquid, solutions, and gases?
How will this affect the total pressure of solids,
liquid, and gases?
How will it affect the value of K?
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Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume increases the
concentration of all gases, but not solids, liquid, or
solutions
same number of moles, but different number of liters, resulting
in a different molarity
• Decreasing the container volume will increase the total
pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the
gases will increase – Dalton’s Law of Partial Pressures
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules
shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same
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The Effect of Volume Changes on Equilibrium
Because there are more
When
the pressure
is
gas
molecules
on the
decreased
by side
increasing
reactants
of the the
volume,
the position
of
reaction,
when
the pressure
equilibrium
toward the
is increasedshifts
the position
of
side with
the greater
equilibrium
shifts
toward the
number
molecules
– the
side withoffewer
molecules
reactant
to decrease
theside
pressure
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The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
•
endothermic reactions absorb energy
If we write Heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, it will help us use Le Châtelier’s
Principle to predict the effect of temperature
changes
even though heat is not matter and not written in a
proper equation
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The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants
Adding heat to an exothermic reaction will
decrease the value of K
How will decreasing the temperature affect the
system?
aA + bB cC + dD + Heat
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
•
•
•
equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products
Adding heat to an endothermic reaction will
increase the value of K
How will decreasing the temperature affect the
system?
Heat + aA + bB cC + dD
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Kc
c
d
a
b
C D
A B
Copyright 2011 Pearson Education, Inc.
The Effect of Temperature Changes
on Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more efficient
mechanism
• Catalysts work for both forward and reverse
reactions
• Catalysts affect the rate of the forward and
reverse reactions by the same factor
• Therefore, catalysts do not affect the
position of equilibrium
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Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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Copyright 2011 Pearson Education, Inc.
Practice – Le Châtelier’s Principle
• The reaction 2 SO2(g) + O2(g) 2 SO3(g) with
DH° = −198 kJ is at equilibrium. How will each of
the following changes affect the equilibrium
concentrations of each gas once equilibrium is
restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
Tro: Chemistry: A Molecular Approach, 2/e
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shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect
Copyright 2011 Pearson Education, Inc.