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Chapter 15
Overview
• Equilibrium Reactions
• Equilibrium Constants Kc & Kp
• Equilibrium Expression product/reactant
• Reaction Quotient
Q
• Calculations
• Le Chatelier’s Principle
– disturbing the equilibrium
Overview, cont’d
• All reactions are reversible
• Dynamic Equilibrium
– When the rates of the forward and
– reverse reactions are equal
• Reactions do not “go to completion”
• Cannot use stoichiometric methods to calculate
amount of products formed
Equilibrium and Rates
A
B
kf [A] = kr [B]
forward rate = reverse rate
[B ]
[A]
= kf = constant
kr
Equilibrium Constant Expression
• Ratio of Products to Reactant
– raised to each coefficient
• for example,
– N2
+
–K =
[NH3]2
[N2][H2]3
3H2
2NH3
cont’d
N2
+
3H2
2NH3
Large K, more product
Product Favored
Small K, more reactant
Reactant Favored
In General:
aA + bB
cC + dD
products
Kc = [C]c[D]d
[A]a[B]b
concentration
reactants
Forms of Eq. Constant Expression
CO(g)
+
H2O(g)
CO2(g)
+
H2(g)
Kc = [CO2][H2]
[CO][H2O]
4HCl(g) + O2(g)
Kc = [Cl2]2[H2O]2
[O2][HCl]4
2H2O(g) + 2Cl2(g)
Cont’d
2HI
H2
+
I2
Kf = [H2][I2]
[HI]2
H2
+
I2
Kr
2HI
=
[HI]2
[H2][I2]
=
1
Kf
Cont’d
2{ 2HI
H2
+
I2 }
Kf = [H2][I2]
[HI]2
4HI
2H2
+
2I2
K = Kf2 = [H2]2[I2]2
[HI]4
Example
N2O4
2NO2
Initial
N2O4
Initial
NO2
0.0
0.0
0.0
0.02
0.02
0.03
0.04
0.0
Kc = [NO2]2
[N2O4]
Equilibrium
N2O4
0.0014
0.0028
0.0045
0.0045
Equilibrium
NO2
Kc
0.0172
0.0243
0.0310
0.0310
0.211
0.211
0.213
0.213
generally unitless
Review values of K:
N2
+
2NH3
3H2
Large K, more product
K>>1
N2
Product Favored
+
3H2
2NH3
Small K, more reactant
Reactant Favored
K<<1
Example:
N2
+
3H2
Kc = 4.34 x 10-3 at 300°C
2NH3
=
[NH3]2
[H2]2 [N2]
• What is K for reverse reaction?
• What is K for 2N2 + 6H2
• What is K for 4NH3
Kc reverse = 230
4NH3 ?
Kc = 1.88 x 10-5
6H2 + 2N2 ?
Kc = 5.31 x 104
Heterogeneous Equilibria:
• When pure solid or liquid is involved
– Pure solids & liquids do not appear in the equilibrium
constant expression
• When H2O is a reactant or product and is the
solvent
– H2O does not appear in the equilibrium constant
expression
Examples:
• CaCO3(s)
CaO(s) + CO2(g)
– conc. = mol = g/cm3 = density
cm3
g/mol
MM
• Kc = [CaO] [CO2] = (constant 1) [CO2]
[CaCO3]
(constant 2)
• Kc’ = Kc (constant 2) = [CO2]
(constant 1)
both are
constant
Multi-Step Equilibria
AgCl(s)
Ag+(aq) +
Ag+(aq) + 2NH3(aq)
AgCl(s) + 2NH3(aq)
Cl-(aq)
K1 = [Ag+][Cl-]
Ag(NH3)2+(aq) K2 = [Ag(NH3)2+]
[Ag+][NH3]2
Ag(NH3)2+(aq) + Cl-(aq)
Ktot = K1 K2 = [Ag(NH3)2+][Cl-]
[NH3]2
Problem:
1.00 mole of H2 & 1.00 mole of I2 are placed in a 1.0 L container at
520 K and allowed to come to equilibrium. Analysis reveals 0.12
mol of HI present at equilibrium. Calculate Kc.
H2(g) +
I2(g)
2HI(g)
initial 1.00
change -0.06
equil.
Kc =
0.94 mol
(0.12)2
=
(0.94)(0.94)
1.00
-0.06
0
+0.12
0.94 mol
+0.12
1.6 x 10-2
Conversion between Kp and Kc
• Kc
– Equilibrium constant using concentrations
• Kp
– Equilibrium constant using partial pressures
Kp = Kc (RT)Dn
P = n RT
V
R = 0.0821 L atm/mol K
T = Temperature in K
Dn = tot. mol product - tot. mol reactant
Problem:
For 2SO3(g)
2SO2(g) + O2(g)
Kc = 4.08 x 10-3 at 1000 K. Calculate Kp.
Kp = Kc(RT)Dn = 4.08 x 10-3 (0.0821 x 1000)1
Kp = 0.0335
Problem:
For 3H2(g) + N2(g)
2NH3(g)
Kc = 0.105 at 472°C. Calculate Kp.
Kp = Kc(RT)Dn = 0.105 (0.0821 x 745)-2
Kp = 2.81 x 10-5
Applications of Eq. Constants
• Reaction Quotient
– Non-equilibrium concentrations used in the
equilibrium constant expression
Q = K Reaction is at equilibrium
Q > K
Reaction will shift left to equilibrium
Q < K
Reaction will shift right to equilibrium
Problem:
Kc = 5.6 x 10-12 at 500 K for
I2(g)
2I(g)
[I2] = 0.020 M & [I] = 2.0 x 10-8 M. Is the reaction at
equilibrium? Which direction will it shift to reach
equilibrium?
Q = [I ]2
[I2]
=
Q
(2.0 x 10-8)2
(0.020)
<
(2.0 x 10-14) <
= 2.0 x 10-14
not at equilibrium
because Q  K
K
(5.6 x 10-12)
Reaction Shifts Right to get to equilibrium
Calculation of Eq. Concentrations
• use the stoichiometry of reaction
• initial concentration of all species
• change that occurs to all species
• equilibrium concentration of all species
• reaction will occur to reach the equilibrium point
no matter the direction of reaction
Problem:
Cyclohexane, C6H12(g), can isomerize to form
methylcyclopentane, C5H9CH3(g). The equilibrium constant at
25°C is 0.12. If the original amount was 0.045 mol cyclohexane
in a 2.8 L flask, what are the concentrations at equilibrium?
C6H12
initial
change
equil.
0.045 mol
-x
0.045 - x
C5H9CH3
0
+x
x
Cont’d
0.12
=
(x)
(0.045 - x)
Solve for x which is the equilibrium concentration
of methylcyclopentane or the product
x = 4.8 x 10-3 mol C5H9CH3 in 2.8 L flask
[C5H9CH3] = 1.7 x 10-3 M
[C6H12] =
1.4 x 10-2 M
Problem:
For the reaction H2 + I2
2HI the Kc = 55.64. You start
with 1.00 mol H2 and 1.00 mol I2 in a 0.500 L flask. Calculate
the equilibrium concentrations of all species?
H2
initial
change
+
I2
2HI
2.00 M
2.00 M
0
-x
-x
+2x
equil. 2.00 - x
2.00 - x
2x
perfect
square
Cont’d
55.64 =
(55.64)½ =
(2x)2
(2.00 -x)2
(2x)2
(2.00 -x)2
½
x = 1.58 M
[HI] = 3.16 M
= 2x
[H2] = 0.42 M
= 2.00 - x
[I2] = 0.42 M
= 2.00 - x
Problem:
For the reaction H2 + I2
2HI the Kc = 55.64. You start with
1.00 mol H2 and 0.50 mol I2 in a 0.500 L flask. Calculate the
equilibrium concentrations of all species?
H2
initial
change
equil.
+
I2
2HI
2.00 M
1.00 M
0
-x
-x
+2x
2.00 - x
1.00 - x
2x
Cont’d
Kc
55.64
=
=
[HI]2
[H2][I2]
not a perfect square
(2x)2
(2.00 -x)(1.00 -x)
reduces to a quadratic equation:
x2 - 3.232 x + 2.155 = 0
Quadratic Equation
x =-b±
b2
- 4ac
2a
½
for
ax2 + bx + c = 0
Cont’d
x = +3.232 ± 10.446 - 8.62
2
½
x = 1.616 ± 0.6755
[HI] = 1.88 M
= 2x
[H2] = 1.06 M
= 2.00 - x
[I2] = 0.060 M
= 1.00 - x
Le Chatelier’s Principle
• When a stress is applied to an equilibrium
reaction, the equilibrium shifts to reduce the
stress
• Types of Stress
– Addition or removal of reactant
– Addition or removal or product
– Increase or decrease of temperature
– Change in pressure or volume
2NOCl(g)
Cl2(g) +
temp.
NOCl
Cl2
NO
temp.
NOCl
Cl2
NO
pressure
volume
2NO(g)
reaction
shifts
reaction
shifts
reaction
shifts
DH = +77 kJ
Addition of Reactant or Product
[C6H12]
[C5H9CH3]
initial 1.4 x 10-2 + 1.0 x 10-2 M
1.7 x 10-3 M
change
-x
equil. 2.4 x 10-2 - x
0.12 = (1.7 x 10-3 + x)
(2.4 x 10-2 - x)
[C6H12] = 0.023 M
+x
1.7 x 10-3 + x
x = 1.05 x 10-3 M
[C5H9CH3] = 0.0028 M
Changes in Temperature
• will change K
• for an endothermic reaction
– increasing T increases K
• for an exothermic reaction
– increasing T decreases K
without
catalyst
Effect of a Catalyst
rr
rf
with
catalyst
Energy
Ea (f)
Ea (r)
A
B
Reactions Path
Reaction Mechanisms & Equilibria
2O3(g)
O3(g)
O(g)
k1
k2
+
3O2(g)
overall rxn
O2(g) + O(g)
O3(g)
k3
fast
equil. rate1 = k1[O3]
rate2 = k2[O2][O]
2O2(g)
slow
rate3 = k3[O][O3]
rate 3 includes the conc. of an intermediate and the
exptl. rate law will include only species that are
present in measurable quantities
Substitution Method
at equilibrium
k1[O3] = k2[O2][O]
rate3 =k3[O][O3]
[O] =
substitute
rate3 = k3k1 [O3]2
k2 [O2]
overall rate
or
= k’ [O3]2
[O2]
k1 [O3]
k2 [O2]
Problem:
Derive the rate law for the following reaction given
the mechanism step below:
OCl -
(aq)
+ I
k1
OCl - + H2O
I-
HOI
k2
k3
+ HOCl
+ OH
-
OI
(aq)
k4
HOCl
+ Cl
(aq)
+
OH Cl -
HOI
+
H2O
+ OI -
(aq)
fast
slow
fast
Cont’d
rate1 = k1 [OCl -][H2O] = rate 2 = k2 [HOCl][OH -]
[HOCl] = k1[OCl -][H2O]
k2[OH -]
rate 3 = k3 [HOCl][I -]
rate 3 = k3k1[OCl -][H2O][I -]
k2 [OH -]
overall rate = k’ [OCl -][I -]
[OH -]
solvent