Transcript Chemical Equilibrium

C h a p t e r 13
Chemical Equilibrium
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
Chemical equilibrium is achieved when:
•
the rates of the forward and reverse reactions are equal and
•
the concentrations of the reactants and products remain
constant
Physical equilibrium
H2O (l)
H2O (g)
Chemical equilibrium
N2O4 (g)
colorless
2NO2 (g)
brown
N2O4 (g)
2NO2 (g)
Demo
equilibrium
equilibrium
equilibrium
Start with NO2
Start with N2O4
Start with NO2 & N2O4
N2O4 (g)
2NO2 (g)
constant
Using Equilibrium Constants
•
01
We can make the following generalizations
concerning the composition of equilibrium mixtures:
If Kc > 103, products predominate over reactants. If Kc is
very large, the reaction is said to proceed to completion.
If Kc is in the range 10–3 to 103, appreciable
concentrations of both reactants and products are present.
If Kc < 10–3, reactants predominate over products. If Kc is
very small, the reaction proceeds hardly at all.
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)
(at 500 K) Kc = 57.0
2H2(g) + O2(g)
2H2O(g)
(at 500 K) Kc = 2.4 x 1047
N2O4 (g)
K=
[NO2]2
[N2O4]
aA + bB
K=
[C]c[D]d
[A]a[B]b
2NO2 (g)
= 4.63 x 10-3
cC + dD
Law of Mass Action
N2O4 (g)
K=
[NO2]2
[N2O4]
2NO2 (g)
= 4.63 x
10-3
2NO2 (g)
N2O4 (g)
[N2O4]
1
= 216
K‘ =
=
2
K
[NO2]
When the equation for a reversible reaction
is written in the opposite direction, the
equilibrium constant becomes the reciprocal
of the original equilibrium constant.
Write the Equilibrium Constant Kc
If you multiply both side of an equilibrium reaction by
n the equilibrium constant should be raised to the
power of n.
N2(g) + 3H2(g)
2NH3(g)
2N2(g) + 6H2(g)
2NH3(g)
N2(g) + 3H2(g)
4NH3(g)
Kc =
Kc´ =
Kc´´=
[NH3]2
[N2][H2]3
[N2][H2]3
[NH3]2
[NH3]4
[N2]2[H2]6
=
1
Kc
= Kc
2
Homogenous equilibrium applies to reactions in which all
reacting species are in the same phase.
N2O4 (g)
Kc =
[NO2
2NO2 (g)
]2
Kp =
[N2O4]
2
PNO
2
PN2O4
In most cases
Kc  Kp
aA (g) + bB (g)
cC (g) + dD (g)
Kp = Kc(RT)Dn
Dn = moles of gaseous products – moles of gaseous reactants
= (c + d) – (a + b)
Homogeneous Equilibrium
CH3COOH (aq) + H2O (l)
[CH3COO-][H3O+]
Kc‘ =
[CH3COOH][H2O]
CH3COO- (aq) + H3O+ (aq)
[H2O] = constant
[CH3COO-][H3O+]
= Kc‘ [H2O]
Kc =
[CH3COOH]
General practice not to include units for the
equilibrium constant.
The concentration of pure liquids are not included
in the expression for the equilibrium constant.
The equilibrium concentrations for the reaction between
carbon monoxide and molecular chlorine to form COCl2 (g)
at 740C are [CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] =
0.14 M. Calculate the equilibrium constants Kc and Kp.
CO (g) + Cl2 (g)
COCl2 (g)
[COCl2]
0.14
=
= 220
Kc =
[CO][Cl2]
0.012 x 0.054
Kp = Kc(RT)Dn
Dn = 1 – 2 = -1
R = 0.0821
T = 273 + 74 = 347 K
Kp = 220 x (0.0821 x 347)-1 = 7.7
The equilibrium constant Kp for the reaction
2NO2 (g)
2NO (g) + O2 (g)
is 158 at 1000K. What is the equilibrium pressure of O2 if
the PNO2 = 0.400 atm and PNO = 0.270 atm?
Kp =
2
PNO
PO2
2
PNO
2
PO2 = Kp
2
PNO
2
2
PNO
PO2 = 158 x (0.400)2/(0.270)2 = 347 atm
Heterogenous equilibrium applies to reactions in which
reactants and products are in different phases.
CaCO3 (s)
[CaO][CO2]
Kc‘ =
[CaCO3]
[CaCO3]
Kc = [CO2] = Kc‘ x
[CaO]
CaO (s) + CO2 (g)
[CaCO3] = constant
[CaO] = constant
Kp = PCO2
The concentration of pure solids and pure liquids are not
included in the expression for the equilibrium constant.
CaCO3 (s)
CaO (s) + CO2 (g)
PCO 2 = Kp
PCO 2 does not depend on the amount of CaCO3 or CaO
Consider the following equilibrium at 295 K:
NH4HS (s)
NH3 (g) + H2S (g)
The partial pressure of each gas is 0.265 atm. Calculate
Kp and Kc for the reaction?
Kp = PNH PH S = 0.265 x 0.265 = 0.0702
3
2
Kp = Kc(RT)Dn
Kc = Kp(RT)-Dn
Dn = 2 – 0 = 2
T = 295 K
Kc = 0.0702 x (0.0821 x 295)-2 = 1.20 x 10-4
Writing Equilibrium Constant Expressions
•
The concentrations of the reacting species in the liquid
phase are expressed in M. In the gaseous phase, the
concentrations can be expressed in M or in atm.
•
The concentrations of pure solids and pure liquids, do not
appear in the equilibrium constant expressions.
•
The equilibrium constant is a dimensionless quantity.
•
In quoting a value for the equilibrium constant, you must
specify the balanced equation and the temperature.
•
If a reaction can be expressed as a sum of two or more
reactions, the equilibrium constant for the overall reaction is
given by the product of the equilibrium constants of the
individual reactions.
Review
N2O4 (g)
2NO2 (g)
Demo
equilibrium
equilibrium
equilibrium
Start with NO2
Start with N2O4
Start with NO2 & N2O4
The reaction quotient (Qc) is calculated by substituting the
initial concentrations of the reactants and products into the
equilibrium constant (Kc) expression.
IF
[NO2]2
Kc =
[N2O4]
[NO2]02
Qc =
[N2O4]0
•
Qc > Kc system proceeds from right to left to reach equilibrium
•
Qc = Kc the system is at equilibrium
•
Qc < Kc system proceeds from left to right to reach equilibrium
Use reaction quotient to predict the direction of shift when the
volume is halved in the following equilibrium:
N2O4(g) æ 2 NO2(g),
•
Consider the reaction: N2O4(g) æ 2 NO2(g), taking place in a
cylinder with a volume = 1 unit.
[NO2] = 2X mol/1 = 2X
[N2 O4 ] = X mol/1 = y-X
K=
[NO2 ]2
[N2 O4 ]
Le Châtelier’s Principle
10
•
The Volume is then halved, which is equivalent to
doubling the pressure.
•
Since Q > K, the [product] is too high and the
reaction progresses in the reverse direction.
At 12800C the equilibrium constant (Kc) for the reaction
Br2 (g)
2Br (g)
Is 1.1 x 10-3. If the initial concentrations are [Br2] = 0.063
M and [Br] = 0.012 M, calculate the concentrations of these
species at equilibrium.
Let (x) be the change in concentration of Br2
Initial (M)
Change (M)
Equilibrium (M)
[Br]2
Kc =
[Br2]
Br2 (g)
2Br (g)
0.063
0.012
-x
+2x
0.063 - x
0.012 + 2x
(0.012 + 2x)2
= 1.1 x 10-3
Kc =
0.063 - x
Solve for x
(0.012 + 2x)2
= 1.1 x 10-3
Kc =
0.063 - x
4x2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x
4x2 + 0.0491x + 0.0000747 = 0
-b ± b2 – 4ac
2
x=
ax + bx + c =0
2a
x = -0.0105 x = -0.00178
Initial (M)
Change (M)
Equilibrium (M)
Br2 (g)
2Br (g)
0.063
0.012
-x
+2x
0.063 - x
0.012 + 2x
At equilibrium, [Br] = 0.012 + 2x = -0.009 M or 0.00844 M
At equilibrium, [Br2] = 0.063 – x = 0.0648 M
Calculating Equilibrium Concentrations
1. Express the equilibrium concentrations of all species in
terms of the initial concentrations and a single unknown x,
which represents the change in concentration.
2. Write the equilibrium constant expression in terms of the
equilibrium concentrations. Knowing the value of the
equilibrium constant, solve for x.
3. Having solved for x, calculate the equilibrium
concentrations of all species.
Predicting the direction of a Reaction
•
The reaction quotient (Qc) is obtained by
substituting initial concentrations into the
equilibrium constant. Predicts reaction direction.
Qc > Kc
System proceeds to form reactants.
Qc = Kc
System is at equilibrium.
Qc < Kc
System proceeds to form products.
Le Châtelier’s Principle
•
Le Châtelier’s principle:
If an external stress
is applied to a system
at equilibrium, the
system adjusts in
such a way that the
stress is partially
offset.
01
Le Châtelier’s Principle
•
02
Concentration Changes:
The concentration stress of an added reactant or
product is relieved by reaction in the direction
that consumes the added substance.
The concentration stress of a removed reactant or
product is relieved by reaction in the direction
that replenishes the removed substance.
Le Châtelier’s Principle: Haber process
N2(g) + 3 H2(g) æ 2 NH3(g)
Exothermic
Cat: iron or
ruthenium
Le Châtelier’s Principle
•
06
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) æ 2 Fe(l) + 3 CO2(g)
•
Use Le Châtelier’s principle to predict the direction
of reaction when an equilibrium mixture is
disturbed by:
•(a)
Adding Fe2O3
(b) Removing CO2 (c) Removing CO
Le Châtelier’s Principle
•
07
Volume and Pressure Changes: Only reactions
containing gases are affected by changes in
volume and pressure.
Increasing pressure = Decreasing volume
•
PV = nRT tells us that increasing pressure or
decreasing volume increases concentration.
Le Châtelier’s Principle
•
N2(g) + 3 H2(g) æ 2 NH3(g)
08
Kc = 0.291 at 700 K
Le Châtelier’s Principle
•
11
Does the number of moles of reaction products
increase, decrease, or remain the same when
each of the following equilibria is subjected to a
decrease in pressure by increasing the volume.
1. PCl5(g)
æ PCl3(g) + Cl2(g)
2. CaO(s)
+ CO2(g) æ CaCO3(s)
3. 3
Fe(s) + 4 H2O(g) æ Fe3O4(s) + 4 H2(g)
Le Châtelier’s Principle
13
•
Temperature Changes: Changes in temperature
can change the equilibrium constant.
•
Endothermic processes
are favored when
temperature increases.
•
Exothermic processes
are favored when
temperature decreases.
Le Châtelier’s Principle
14
Example:
• The reaction N2(g) + 3 H2(g) æ 2 NH3(g) which is
exothermic by 92.2 kJ.
•
Le Châtelier’s Principle
•
15
In the first step of the Ostwald process for synthesis
of nitric acid, ammonia is oxidized to nitric oxide by
the reaction:
4 NH3(g) + 5 O2(g) æ 4 NO(g) + 6 H2O(g) ∆H° = –905.6 kJ
•
How does the equilibrium amount vary with an
increase in temperature?
Le Châtelier’s Principle
16
•
The following pictures represent the composition of
the equilibrium mixture at 400 K and 500 K for the
reaction A(g) + B(g) æ AB(g).
•
Is the reaction
endothermic or
exothermic?
Effect of Catalysis, Reduction of
Activation Energy : No effect