Transcript Slide 1

The Equilibrium State
Chemical Equilibrium: The state reached when the
concentrations of reactants and products remain constant over
time.
N2O4(g)
Colorless
2NO2(g)
Brown
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Chapter
13/1
The Equilibrium State
N2O4(g)
2NO2(g)
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Chapter
13/2
The Equilibrium Constant Kc
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Chapter
13/4
The Equilibrium Constant Kc
For a general reversible reaction:
aA + bB
cC + dD
Double arrows show reversible reaction
Equilibrium equation:
[C]c[D]d
Kc =
[A]a[B]b
Equilibrium constant
Products
Reactants
Equilibrium constant expression
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Chapter
13/5
The Equilibrium Constant Kc
N2O4(g)
2NO2(g)
For the following reaction:
Kc =
[NO2]2
[N2O4]
= 4.64 x 10-3 (at 25 °C)
Examples
 Write the equilibrium constant, Kc, for each of the following
reactions:
a. CH4(g) + H2O(g)
b.
2 NO(g)
CO(g) + 3 H2(g)
N2(g) + O2(g)
The Equilibrium Constant Kc
Learning Check
Experiment 2
Kc =
[NO2]2
[N2O4]
Experiment 5
(0.0125)2
=
0.0337
(0.0141)2
= 4.64 x 10-3
0.0429
= 4.63 x 10-3
The Equilibrium Constant Kc
The equilibrium constant and the equilibrium constant expression are for
the chemical equation as written.
N2(g) + 3H2(g)
2NH3(g)
2NH3(g)
N2(g) + 3H2(g)
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Kc =
Kc´ =
[NH3]2
[N2][H2]3
[N2][H2]3
[NH3]2
=
1
Kc
Chapter
13/10
Equilibrium Constant, Kc
2N2(g) + 6H2(g)
4NH3(g)
K c´´ =
[NH3]4
[N2]2[H2]6
= Kc
2
Examples
 Consider the following equilibria
2CO2(g) + 2 CF4(g)
4COF2(g)
Write the equilibrium constant K’c expresion
The Equilibrium Constant Kp
N2O4(g)
Kp =
P NO
2
PN O
2 4
2NO2(g)
2
P is the partial pressure of that component
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Chapter
13/13
The Equilibrium Constant Kp
Kp =
Kc(RT)Dn
R is the gas constant,
L atm
0.082058
K mol
T is the absolute temperature (Kelvin).
n is the number of moles of gaseous products
minus the number of moles of gaseous
reactants.
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Chapter
13/14
Heterogeneous Equilibria
CaCO3(s)
CaO(s) + CO2(g)
Limestone
Kc =
[CaO][CO2]
[CaCO3]
Lime
=
(1)[CO2]
(1)
= [CO2]
Pure solids and pure liquids are not included.
Kc = [CO2]
Kp = P CO
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2
Chapter
13/15
Examples
 In the industrial synthesis of hydrogen, mixtures of CO and
H2 are enriched in H2 by allowing the CO to react with
steam. The chemical equation for this so-called water-gas
shift reaction is
CO(g) + H2O(g)
CO2(g) + H2(g)
What is the value of Kp at 700K if the partial pressures in an
equilibrium mixture at 700K are 1.31 atm of CO, 10.0 atm
of H2O, 6.12 atm of CO2, and 20.3 atm of H2?
Examples
 Consider the following unbalanced reaction
(NH4)2S(s)
NH3(g) + H2S(g)
An equilibrium mixture of this mixture at a certain
temperature was found to have [NH3] = 0.278 M and [H2S]
= 0.355 M. What is the value of the equilibrium constant
(Kc) at this temperature?
Examples
 Consider the following reaction and corresponding value of
Kc:
H2(g) + I2(s)
2HI(g)
Kc = 6.2 x 102 at 25oC
What is the value of Kp at this temperature?
Heterogeneous Equilibria
Using the Equilibrium Constant
2H2O(g)
(at 500 K)
2H2(g) + O2(g)
Kc = 4.2 x 10-48
H2(g) + I2(g)
2HI(g)
2H2(g) + O2(g)
(at 500 K)
Kc = 57.0
(at 500 K)
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2H2O(g)
Kc = 2.4 x 1047
Chapter
13/20
Using the Equilibrium Constant
aA + bB
Reaction quotient:
cC + dD
Qc =
[C]tc[D]td
[A]ta[B]tb
The reaction quotient, Qc, is defined in the same way as the
equilibrium constant, Kc, except that the concentrations in Qc are
not necessarily equilibrium values.
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Chapter
13/21
Using the Equilibrium Constant
net reaction goes from left to right
(reactants to products).
•
If Qc < Kc
•
If Qc > Kc
net reaction goes from right to left
(products to reactants).
•
If Qc = Kc
no net reaction occurs.
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Chapter
13/22
Finding Equilibrium concentrations from
Initial Concentrations
 Steps to follow in calculating equilibirium concentrations from
initial concentration
 Write a balance equation for the reaction
 Make an ICE (Initial, Change, Equilibrium) table, involves
 The initial concentrations
 The change in concentration on going to equilibrium, defined as x
 The equilibrium concentration
 Substitute the equilibrium concentrations into the equilibrium
equation for the reaction and solve for x
 Calculate the equilibrium concentrations form the calculated value
of x
 Check your answers
Using the Equilibrium Constant
 Sometimes you must use quadratic equation to solve for x,
choose the mathematical solution that makes chemical sense
Quadratic equation
ax2 + bx + c = 0
Finding Equilibrium concentrations
from Initial Concentrations
For Homogeneous Mixture
aA(g)
Initial concentration (M)
[A]initial
Change (M)
- ax
Equilibrium (M)
Kc =
[A]initial – ax
[products]
[reactants]
bB(g)
[B]initial
+ bx
[B]initial – bx
+
cC(g)
[C]initial
+ cx
[C]initial - cx
Finding Equilibrium concentrations
from Initial Concentrations
For Heterogenous Mixture
aA(g)
bB(g)
Initial concentration (M)
[A]initial
[B]initial
Change (M)
- ax
Equilibrium (M)
Kc =
[A]initial – ax
[products]
[reactants]
+ bx
[B]initial – bx
+
cC(s)
……..
……..
……….
Using the Equilibrium Constant
At 700 K, 0.500 mol of HI is added to a 2.00 L container and
allowed to come to equilibrium. Calculate the equilibrium
concentrations of H2, I2, and HI . Kc is 57.0 at 700 K.
H2(g) + I2(g)
2HI(g)
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Chapter
13/27
Using the Equilibrium Constant
Set up a table:
H2(g) + I2(g)
2HI(g)
I
0
0
0.250
C
+x
+x
-2x
E
x
x
0.250 - 2x
Substitute values into the equilibrium expression:
[HI]2
Kc =
[H2][I2]
(0.250 - 2x)2
57.0 =
x2
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Chapter
13/28
Using the Equilibrium Constant
Solve for “x”:
(0.250 - 2x)2
57.0 =
x = 0.0262
x2
Determine the equilibrium concentrations:
H2: 0.0262 M
I2: 0.0262 M
HI: 0.250 - 2(0.0262) = 0.198 M
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Chapter
13/29
Examples
 Consider the following reaction
I2(g) + Cl2(g)
2ICl(g)
Kp = 81.9 (at 25oC)
A reaction mixture at 25oC intially contains PI2 = 0.100 atm,
PCl2 = 0.100 atm, and PICl = 0.100 atm. Find the
equilibrium pressure of I2, Cl2 and ICl at this temperature.
In which direction does the reaction favored?
Examples
 The value of Kc for the reaction is 3.0 x 10-2. Determine
the equilibrium concentration if the initial concentration of
water is 8.75 M
C(s) + H2O(g)
CO(g) + H2(g)
Examples
 Consider the following reaction
N2O4(g)
2 NO2(g)
A reaction mixture at 100oC initially contains [NO2] =
0.100M. Find the equilibrium concentration of NO2 and
N2O4 at this temperature
Le Châtelier’s Principle
Le Châtelier’s Principle: If a stress is applied to a reaction
mixture at equilibrium, net reaction occurs in the direction that
relieves the stress.
•
The concentration of reactants or products can be
changed.
•
The pressure and volume can be changed.
•
The temperature can be changed.
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Chapter
13/33
Altering an Equilibrium Mixture:
Concentration
N2(g) + 3H2(g)
2NH3(g)
John E. McMurry • Robert C. Fay
C H E M I
S T R Y
Fifth Edition
Chapter 13
Chemical Equilibrium
Lecture Notes
Alan D. Earhart
Southeast Community College • Lincoln, NE
Copyright © 2008 Pearson Prentice Hall, Inc.
Altering an Equilibrium Mixture:
Concentration
In general, when an equilibrium is disturbed by the addition or
removal of any reactant or product, Le Châtelier’s principle predicts
that
•
the concentration stress of an added reactant or product is
relieved by net reaction in the direction that consumes the added
substance.
•
the concentration stress of a removed reactant or product is
relieved by net reaction in the direction that replenishes the
removed substance.
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Chapter
13/36
Altering an Equilibrium
Mixture: Concentration
 Add reactant – denominator in Qc expression becomes
larger
 Qc < Kc
 To return to equilibrium, Qc must be increases
 More product must be made => reaction shifts to the right
Altering an Equilibrium
Mixture: Concentration
 Remove reactant – denominator in Qc expression becomes
smaller
 Qc > Kc
 To return to equilibrium, Qc must be decreases
 Less product must be made => reaction shifts to the left
Altering an Equilibrium Mixture:
Concentration
N2(g) + 3H2(g)
2NH3(g)
at 700 K, Kc = 0.291
An equilibrium mixture of 0.50 M N2, 3.00 M H2, and 1.98 M NH3 is
disturbed by increasing the N2 concentration to 1.50 M.
Qc =
[NH3]2
[N2][H2]3
=
(1.98)2
(1.50)(3.00)3
= 0.0968 < Kc
Since Qc < Kc, more reactants will be consumed and the
net reaction will be from left to right.
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Chapter
13/39
Examples
 The reaction of iron (III) oxide with carbon monoxide
occurs in a blast furnace when iron ore is reduced to iron
metal:
Fe2O3(s) + 3 CO(g)
2 Fe(l) + 3CO2(g)
Use Le Chatellier’s principle to predict the direction of net
reaction when an equilibrium mixture is disturbed by:
a. adding Fe2O3
b. Removing CO2
c. Removing CO; also account for the change using the
reaction quotient Qc
Examples
 Consider the following reaction at equilibrium
CO(g) + Cl2(g)
COCl2(g)
Predict whether the reaction will shift left, shift right, or
remain unchanged upon each of the following reaction
mixture
a. COCl2 is added to the reaction mixture
b. Cl2 is added to the reaction mixture
c. COCl2 is removed from the reaction mixture: also
account for the change using the reaction quotient Qc
Altering an Equilibrium Mixture:
Pressure and Volume
N2(g) + 3H2(g)
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2NH3(g)
Chapter
13/42
Altering an Equilibrium Mixture:
Pressure and Volume
In general, when an equilibrium is disturbed by a change in volume
which results in a corresponding change in pressure, Le Châtelier’s
principle predicts that
•
an increase in pressure by reducing the volume will bring
about net reaction in the direction that decreases the number
of moles of gas.
•
a decrease in pressure by enlarging the volume will bring
about net reaction in the direction that increases the number
of moles of gas.
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Chapter
13/43
Altering an Equilibrium Mixture:
Pressure and Volume
 If reactant side has more moles of gas
 Denominator will be larger



Qc < Kc
To return to equilibrium, Qc must be increased
Reaction shifts toward fewer moles of gas (to the product)
 If product side has more moles of gas
 Numerator will be larger



Qc > Kc
To return to equilibrium, Qc must be decreases
Reaction shifts toward fewer moles of gas (to the reactant)
Altering an Equilibrium Mixture:
Pressure and Volume
 Reaction involves no change in the number moles of gas
 No effect on composition of equilibrium mixture
 For heterogenous equilibrium mixture
 Effect of pressure changes on solids and liquids can be
ignored

Volume is nearly independent of pressure
 Change in pressure due to addition of inert gas
 No change in the molar concentration of reactants or
products
 No effect on composition
Examples
 Consider the following reaction at chemical equilibrium
2 KClO3(s)
2 KCl(s) + 3O2(g)
a. What is the effect of decreasing the volume of the
reaction mixture?
b. Increasing the volume of the reaction mixture?
c. Adding inert gas at constant volume?
Examples
 Does the number moles of products increases, decreases or
remain the same when each of the following equilibria is
subjected to a increase in pressure by decreasing the
volume?
 PCl5(g)
PCl3(g) + Cl2(g)
 CaO(s) + CO2(g)
CaCO3(s)
 3 Fe(s) + 4H2O(g)
Fe3O4(s) + 4 H2(g)
Altering an Equilibrium Mixture:
Temperature
N2(g) + 3H2(g)
2NH3(g)
DH° = -2043 kJ
As the temperature increases, the equilibrium shifts from
products to reactants.
Altering an Equilibrium Mixture:
Temperature
In general, when an equilibrium is disturbed by a change in
temperature, Le Châtelier’s principle predicts that
•
•
the equilibrium constant for an exothermic reaction (negative
DH°) decreases as the temperature increases.
• Contains more reactant than product
• Kc decreases with increasing temperature
the equilibrium constant for an endothermic reaction
(positive DH°) increases as the temperature increases.
• Contains more product than reactant
• Kc increases with increasing temperature
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Chapter
13/49
Examples
 The following reaction is endothermic
CaCO3(s)
CaO(s) + CO2(g)
a. What is the effect of increasing the temperature of the
reaction mixture?
b. Decreasing the temperature?
Examples
 In the first step of Ostwald process for the synthesis of
nitric acid, ammonia is oxidized to nitric oxide by the
reaction:
2 NH3(g) + 5 O2(g)
4 NO(g) + 6 H2O(l)
ΔHo = -905 kJ
How does the temperature amount of NO vary with an
increases in temperature?
The Effect of a Catalyst on
Equilibrium
The Effect of a Catalyst on
Equilibrium
 Catalyst increases the rate of a chemical reaction
 Provide a new, lower energy pathway
 Forward and reverse reactions pass through the same
transition state
 Rate for forward and reverse reactions increase by the same
factor
 Does not affect the composition of the equilibrium mixture
 Does not appear in the balance chemical equation
 Can influence choice of optimum condition for a reaction