Chemical Kinetics &
Equilibrium
Chapter 16
Collision Model
Key Idea: Molecules must collide to react.
However, only a small fraction of collisions
produces a reaction. Why?
For a reaction to occur, molecules must
have:
1. sufficient energy to break old bonds.
2. proper orientation for collision to be
effective.
Three possible collision orientations-- a) & b) produce reactions,
while c) does not.
Factors Affecting Rate of
Reaction
Concentration: The higher the concentration
of the reactants, the more likely an
effective collision will occur.
Temperature: An increase in temperature
increases:
1. the energy of a collision.
2. the number of collisions.
12_300
T1
T2 > T1
T2
0
0
Ea
Energy
Plot showing the number of collisions with a particular
energy at T1& T2, where T2 > T1 -- Boltzman Distribution.
Activation Energy, Ea
Activation energy for a given reaction is a
constant and not temperature dependent.
Activation energy represents the minimum
energy for a reaction to occur.
a) The change in potential energy as a function of reaction progress.
Ea is the activation energy and E is the net energy change -exothermic. b) Molecular representation of the reaction.
Enthalpy -- H
Enthalpy -- at constant pressure, the change in
enthalpy equals the energy flow as heat.
Exothermic -- H is negative (-).
Endothermic -- H is positive (+).
Catalysis
Catalyst: A substance that speeds up a reaction
without being consumed
Enzyme: A large molecule (usually a protein)
that catalyzes biological reactions.
Homogeneous catalyst: Present in the same
phase as the reacting molecules.
Heterogeneous catalyst: Present in a different
phase than the reacting molecules.
12_303
Uncatalyzed
pathway
Energy
Catalyzed
pathway
Products
E
Reactants
Reaction progress
Energy plots for a catalyzed and an uncatalyzed pathway
for an endothermic reaction.
Number of collisions
with a given energy
Effective
collisions
(uncatalyzed)
Number of collisions
with a given energy
12_304
Ea (catalyzed )
E a (uncatalyzed )
Energy
(a)
Effective
collisions
(catalyzed)
Energy
(b)
Effect of a catalyst on the number of reaction-producing
collisions. A greater fraction of collisions are effective
for the catalyzed reaction.
Catalysis
The breakdown of the ozone layer is illustrated by
the following equation:
Cl + O3 ---> ClO + O2
O + ClO ---> Cl + O2
Cl + O3 + O + ClO ---> ClO + O2 + Cl + O2
Cl
Net Reaction: O + O3 ----> 2O2
What is the catalyst? The intermediate?
Chemical Equilibrium
The state where the concentrations of all
reactants and products remain constant with
time.
On the molecular level, there is frantic
activity. Equilibrium is not static, but is a
highly dynamic situation.
Reactions That Appear to Run to
Completion
1. Formation of a precipitate.
2. Formation of a gas.
3. Formation of a molecular substance such
as water.
These reactions appear to run to completion,
but actually the equilibrium lies very far to
the right. All reactions in closed vessels
reach equilibrium.
13_1577
(a)
(b)
(c)
(d)
Molecular representation of the reaction 2NO2(g) ---->
N2O4(g). c) & d) represent equilibrium.
Figure 16.9: (a) The initial equilibrium mixture of N2,
H2 and NH3. (b) Addition of N2. (c) The new
equilibrium.
Chemical Equilibrium
2NO2(g) <----> N2O4(g)
The forward reaction goes to the right.
The reverse reaction goes to the left.
At equilibrium the rate of the reverse reaction
equals the rate of the forward reaction.
13_315
Concentration
Equilibrium
H2
NH3
N2
Time
Concentration profile for the Haber Process which
begins with only H2(g) & N2(g).
The Law of Mass Action
For
jA + kB  lC + mD
The law of mass action (Cato Guldberg &
Peter Waage) is represented by the
equilibrium expression:
l
m
C D
K
j
k
A B
Equilibrium Expression
l
m
C D
K
j
k
A B
K is the equilibrium constant.
[C] is the concentration expressed in mol/L.
K is temperature dependent.
Equilibrium Constant, K
For an exothermic reaction, if the temperature
increases, K decreases.
For an endothermic reaction, if the
temperature increases, K increases.
Writing Equilibrium
Expressions
2O3(g) <---> 3O2(g)
O 

K
O 
3
2
2
3
Writing Equilibrium Expressions
Write the equilibrium expression for the
following:
H2(g) + F2(g) <---> 2HF(g)
K
 HF 2
H  F 
2
2
NH 

K
N H 
2
N2(g) + 3H2(g) <---> 2NH3(g)
3
3
2
2
Equilibrium Expression
Write the equilibrium expression for the
following reaction:
4NH3(g) + 7O2(g)  4NO2(g) + 6H2O(g)
4
6
NO2 H2O
K
4
7
NH3 O2
Table 16.1: Results of Three Experiments for the
Reaction N2(g) 1 3H2(g)
2NH3(g) at 500 ºC
Equilibrium Position
For a given reaction at a given temperature,
there is only one equilibrium constant (K),
but there are an infinite number of
equilibrium positions.
Where the equilibrium position lies is
determined by the initial concentrations of
the reactants and products. The initial
concentrations do not affect the
equilibrium constant.
Homogeneous Equilibria
Homogeneous equilibria are equilibria in
which all substances are in the same state.
N2(g) + 3H2(g) <---> 2NH3(g)
H2(g) + F2(g) <---> 2HF(g)
Heterogeneous Equilibria
. . . are equilibria that involve more than one
phase.
CaCO3(s)  CaO(s) + CO2(g)
K = [CO2]
The position of a heterogeneous equilibrium
does not depend on the amounts of pure
solids or liquids present. This does not apply
to gases or solutions.
13_1579
CaCO3
CaO
CaCO3
CaO
The position of the equilibrium CaCO3(s) ---> CaO(s) + CO2(g)
does not depend upon the amounts of solid CaCO3 or CaO.
Figure 16.10: The reaction system
CaCO3(s) --> CaO(s) + CO2(g)
Heterogeneous Equilibria
Write equilibrium expressions for the following:
2HOH(l) <---> 2H2(g) + O2(g)
K = [H2]2[O2]
2HOH(g) <---> 2H2(g) + O2(g)
H  [O ]

K
H O
2
2
2
2
2
Heterogeneous Equilibria
Write equilibrium expressions for the following:
PCl5(s) <---> PCl3(l) + Cl2(g)
K = [Cl2]
CuSO4. 5 HOH(s) <---> CuSO4(s) + 5HOH(g)
K = [HOH]5
Le Châtelier’s Principle
. . . If a system at equilibrium is
subjected to a stress, the
equilibrium will be displaced in
such direction as to relieve the
stress.
Le Chatelier’s Principle
If a reactant or product is added to a system at
equilibrium, the system will shift away
from the added component.
If a reactant or product is removed, the system
will shift toward the removed component.
Effect of Changes in
Concentration on Equilibrium
N2(g) + 3H2(g) <---> 2NH3(g)
The addition of 1.000 M N2 has the following effect:
Equilibrium Position I
Equilibrium Position II
[N2] = 0.399M
[N2] = 1.348 M
[H2] = 1.197 M
[H2] = 1.044 M
[NH3] = 0.203 M
[NH3] = 0.304 M
Effect of Change in Concentration on
Equilibrium
NH 
N H 
2
Position I
K
3
3
2
2
 0.203 2
K
3
 0.3991.197
K = 0.0602
Position II
NH 
N H 
2
K
3
3
2
2
 0.304 2
K
1.3481.044 3
K = 0.0602
Changes in Concentration
Predict the effect of the changes listed to this
equilbrium:
As4O6(s) + 6C(s) <---> As4(g) + 6CO(g)
a) addition of carbon monoxide
b) addition or removal of C(s) or As4O6(s)
a) left
b) none
c) removal of As4(g)
c) right
The Effect of Container
Volume on Equilibrium
If the size of a container is changed, the
concentration of the gases change.
A smaller container shifts the equilibrium to
the right -- N2(g) + 3H2(g) ---> 2NH3(g).
Four gaseous molecules produce two
gaseous molecules.
A larger container shifts to the left -- two
gaseous molecules produce four gaseous
molecules.
13_318
Key:
N2
H2
NH3
The system of N2, H2, and NH3 are initially at
equilibrium. When the volume is decreased, the
system shifts to the right -- toward fewer molecules.
The Effect of Container
Volume on Equilibrium
Predict the direction of shift for the following
equilibrium systems when the volume is
reduced:
a) P4(s) + 6Cl2(g) <---> 4PCl3(l)
a) right
b) PCl3(g) + Cl2(g) <---> PCl5(g)
b) right
c) PCl3(g) + 3NH3(g) <---> P(NH2)3(g) + 3HCl(g)
c) none
Equilibrium Constant, K
For an exothermic reaction, if the temperature
increases, K decreases.
N2(g) + 3H2(g) <---> 2NH3(g) + 92 kJ
For an endothermic reaction, if the
temperature increases, K increases.
CaCO3(s) + 556 kJ <---> CaO(s) + CO2(g)
Effect of Temperature on
Equilbirum
Predict how the equilibrium will shift as the
temperature is increased.
N2(g) + O2(g) <---> 2NO(g) (endothermic)
shift to the right
2SO2(g) + O2(g) <---> 2SO3(g) (exothermic)
shift to the left
Effects of Changes on the
System
1. Concentration: The system will shift away
from the added component.
2. Temperature: K will change depending
upon the temperature [treat heat as a
reactant (endothermic) and as a product
(exothermic)].
Effects of Changes on the
System (continued)
3. Pressure:
a. Addition of inert gas does not affect the
equilibrium position.
b. Decreasing the volume shifts the
equilibrium toward the side with fewer
gaseous molecules.
Figure 16.12: Shifting equilibrium by changing the
temperature
Magnitude of K
A K value much larger than 1 means that the
equilibrium system contains mostly
products -- equilibrium lies far to the right.
A very small K value means the system
contains mostly reactants -- equilibrium
lies far to the left.
The size of K and the time required to reach
equilibrium are not directly related!
Calculating Equilibrium
Concentrations
Gaseous phosphorus pentachloride
decomposes to chlorine gas and gaseous
phosphorus trichloride. If K = 8.96 x 10-2 ,
and the equilbrium concentration of PCl5
is 6.70 x 10-3 M and that of PCl3 is 0.300
M, calculate the concentration of Cl2 at
equilibrium.
PCl5(g) <---> PCl3(g) + Cl2(g)
Calculating Equilibrium
Concentrations
PCl5(g) <---> PCl3(g) + Cl2(g)
K = 8.96 x 10-2
3
[PCl5] = 6.70 x 10-3 M
[PCl3] = 0.300 M
PCl  [Cl ]

K
PCl 
K PCl 
[Cl ] 
PCl 
2
5
5
2
[Cl2] = ?
3
[Cl 2 ] 
8.96 x 10-2  6.70 x 10-3 
 0.300
[Cl2] = 2.00 x 10-3 M
Solubility
•
Allows us to flavor foods -- salt & sugar.
•
Solubility of tooth enamel in acids.
•
Allows use of toxic barium sulfate for
intestinal x-rays.
Solubility Product
For solids dissolving to form aqueous solutions.
Bi2S3(s)  2Bi3+(aq) + 3S2(aq)
Ksp = solubility product constant
and
Ksp = [Bi3+]2[S2]3
Why is Bi2S3(s) not included in the solubilty
product expression?
Writing Solubility Product
Expressions
Write the balanced equation for dissolving each of the
following solids in water. Also write the Ksp expression
for each solid.
a) PbCl2(s)
b) Bi2S3(s)
c) Ag2CrO4(s)
PbCl2(s) <---> Pb2+(aq) + 2Cl-(aq) Ksp = [Pb2+][Cl-]2
Bi2S3(s) <---> 2Bi3+(aq) + 3S2-(aq) Ksp = [Bi3+]2[S2-]3
Ag2CrO4(s) <---> 2Ag+(aq) + CrO42-(aq) Ksp = [Ag+]2[CrO42-]
Solubility Product Calculations
Cupric iodate has a measured solubility of 3.3 x
10-3 mol/L. What is its solubility product?
Cu(IO3)2(s) <---> Cu2+(aq) + 2 IO3-(aq)
3.3 x 10-3 M ---> 3.3 x 10-3 M + 6.6 x 10-3 M
Ksp = [Cu2+][IO3-]2
Ksp = [3.3 x 10-3][6.6 x 10-3]2
Ksp = 1.4 x 10-7
Calculating Solubility from Ksp
Values
The Ksp value for solid AgI(s) is 1.5 x 10-16 at 25 oC.
Calculate the solubility of AgI(s) in water at 25 oC.
AgI(s) <---> Ag+(aq) + I-(aq)
[Ag+] = x
Ksp = 1.5 x 10-16 = [Ag+][I-]
[I-] = x
Ksp = 1.5 x 10-16 = x2
x = 1.5 x 10-16
x = 1.2 x 10-8 M
Calculating Solubility from Ksp
Values
The Ksp value for solid lead chromate is 2.0 x 10-16 at 25
oC. Calculate its solubility in water at 25 oC.
PbCrO4(s) <---> Pb2+(aq) + CrO42-(aq)
[Pb2+] = x
]
Ksp = 2.0 x 10-16 = [Pb2+][CrO42-
[CrO42-] = x
Ksp = 2.0 x 10-16 = x2
x = 2.0 x 10-16
-8
Solubility Product Calculations
Cu(IO3)2(s) <---> Cu2+(aq) + 2 IO3-(aq)
Ksp = [Cu2+][IO3-]2
If solid cupric iodate is dissolved in HOH;
double & square the iodate concentration.
If mixing two solutions, one containingCu2+
and and the other IO3-, then use the
concentration of iodate and only square it.