Chemistry 232
Chemical Kinetics
Chemical Kinetics


Chemical kinetics - speed or rate at
which a reaction occurs
How are rates of reactions affected by
• Reactant concentration?
• Temperature?
• Reactant states?
• Catalysts?
The Instantaneous Reaction
Rate


Consider the following reaction
A+BC
Define the instantaneous rate of
consumption of reactant A, A
A
d A 

dt
Reaction Rates and Reaction
Stoichiometry

Look at the reaction
O3(g) + NO(g)  NO2(g) + O2(g)
d O 3 
d[NO]
d[ NO 2 ]
d[ O 2 ]
rate = ==+
=+
dt
dt
dt
dt
Another Example
2 NOCl (g)  2 NO + 1 Cl2 (g)
1 d NOCl 
1 d[NO]
d[Cl 2 ]
rate = =
=+
2 dt
2 dt
dt
WHY? 2 moles of NOCl disappear
for every 1 mole Cl2 formed.
The General Case
aA+bBcC+dD
rate =

-1 d[A] = -1 d[B] = +1 d[C] = +1 d[D]
a dt
b dt
c dt
d dt
Why do we define our rate in this way?
removes ambiguity in the measurement of
reaction rates
Obtain a single rate for the entire equation, not
just for the change in a single reactant or product.
Alternative Definition of the
Rate

Rate of conversion related to the
advancement of the reaction, .
1 d
rate of reaction =
V dt
V = solution volume
An Example

Examine the following reaction
2 N2O5 (g)  4 NO2 (g) + O2 (g)
Initial
Change
Final
N2O5
nø
NO2
0
O2
0
-2
+4
+
nø - 2
4

The N2O5 Decomposition
1 d
rate of reaction =
V dt
Note – constant
volume system
d N 2 O 5 
d  V 
=- 2
dt
dt
d[NO 2 ]
d  V 
= 4
dt
dt
d[O 2 ]
d  V 
=
dt
dt
The Rate Law


Relates rate of the reaction to the
reactant concentrations and rate
constant
For a general reaction
aA+bB+cC  dD+eE
1 d
Rate  v 
 k[A] x [B] y [C] z
V dt
Rate Laws (Cont’d)


The only way that we can determine
the superscripts (x, y, and z) for a nonelementary chemical reaction is by
experimentation.
Use the isolation method (see first year
textbooks).

For a general reaction
 x + y + z = reaction order
e.g. X = 1; Y = 1; Z = 0
2nd order reaction (x + y + z = 2)
X = 0; Y = 0; Z = 1 (1st order reaction)
X = 2; Y = 0; Z = 0 (2nd order)
Rate Laws for Multistep
Processes


Chemical reactions generally proceed
via a large number of elementary steps
- the reaction mechanism
The slowest elementary step  the
rate determining step (rds).
An Example Reaction
Mechanism
O3
k
1 



k
1
O2  O
k2
O3  O 
 2O2

O atoms are intermediates in the above
reaction sequence.
• Intermediates – generally small, indeterminate
concentrations.
Integrated Rate Laws


The rate law gives us information about
how the concentration of the reactant
varies with time
How much reactant remains after
specified period of time? Use the
integrated rate laws.
First Order Reaction
A  product
Rate = v = - d[A]/d t = k[A]
 How does the concentration of the
reactant depend on time?
A
ln
 Ao

  kt

k has units of s-1
The Half-life of a First Order
Reaction

For a first order reaction, the half-life
t1/2 is calculated as follows.
t1 /2
0.693

k
Radioactive Decay



Radioactive Samples decay according
to first order kinetics.
This is the half-life of samples
containing e.g. 14C , 239Pu, 99Tc.
Example
14
14
0
C  N  1 
Second Order Reaction


A  products v = k[A]2
A + B  products v = k[A][B]
Case 1 is 2nd order in A
Case 2 is 1st order in A and B and 2nd
order overall
The Dependence of
Concentration on Time

For a second order process where
v = k[A]2
1
1

 kt
A  A o
Half-life for This Second Order
Reaction.

[A] at t = t½ = ½ [A]0
1
1
=
+ kt 1 / 2
[A ]0 /2 [A ]o
1
or t 1 / 2 =
k [A ]0
Other Second Order Reactions

Examine the Case 2 from above
A + B  products
v = k[A][B]
 B 

B o

ln


A

A o



  kt B o  A o 


A Pseudo-first Order Reaction

Example hydration of methyl iodide
CH3I(aq) + H2O(l)  CH3OH(aq) +
H+(aq) + I-(aq)
Rate = k [CH3I] [H2O]

Since we carry out the reaction in
aqueous solution
[H2O] >>>> [CH3I] \ [H2O] doesn’t
change by a lot
Pseudo-first Order (cont’d)

Since the concentration of H2O is
essentially constant
v = k [CH3I] [constant]
= k` [CH3I] where k` = k [H2O]

The reaction is pseudo first order since
it appears to be first order, but it is
actually a second order process.
Sequential First Order Reactions

Suppose we have two first order reactions
occurring in sequence.
A B
B C
v b  k a  A 
vc  k b  A 
What is Our Intermediate?

B is an intermediate in the above
reaction sequence.
• Clearly B is formed in the first elementary
step of the reaction mechanism and
consumed in the final step.
The Concentration
Dependencies of the Species

The amounts of the reactants are related
to the reaction rates as follows.
d  A 
d C 
 k a  A  ;
 k b B 
dt
dt
d B 
 k a  A   k b B 
dt
Sequential Reactions (3)

For a set of initial conditions [A]o  0, and
[B]o = [C]o =0 mol/L.
 A    A O e
ka t
d  A 
ka t
 k a  A O e
 k b B 
dt


Sequential Reactions (4)

The concentration of the intermediate can be
written
ka
ka t
k b t
e
e
B  
 A O
kb  ka


Sequential Reactions (5)
Note  A o   A   B   C 
Hence C    A o   A   B 
 k a e kb t  k b e ka t

 1   A O
C   
kb  ka


The Rate Determining Step of
the Reaction

What happens if one of the steps in the
reaction is much slower than the other
reaction step?
kb t
ka t




k
e

k
e
 a

b
lim C   lim 
 1   A O 
k b 
k b 
kb  ka






 1e
ka t
  A 
O
Note – assuming ka <<< kb
The Rate Determining Step of
the Reaction (2)

If kb <<< ka
kb t
ka t




k
e

k
e
 a

b
lim C   lim 
 1   A O 
k a 
k a 
kb  ka






 1e
kb t
  A 
O
Temperature Dependence of
Reaction Rates

Reaction rates generally increase with
increasing temperature.

Arrhenius Equation
k  Ae
E a
RT
A = pre-exponential factor
Ea = the activation energy
Reactions Approaching
Equilibrium
Examine the concentration profiles for the
following simple process.
A⇌B
A
Concentration

B
Time
Approaching Equilibrium

Calculate the amounts of A and B at
equilibrium.
v A ,eq



A eq   A o  V  


k
v B ,eq



B eq   V   /


k
The Equilibrium Condition

At equilibrium, vA,eq = vB,eq.
k A eq  v A ,eq  k B eq  v B ,eq
/
B eq
k

K
/
A eq
k
Elementary steps and the
Molecularity

Kinetics of the elementary step depends only
on the number of reactant molecules in that
step!
• Molecularity  the number of reactant
molecules that participate in elementary
steps
The Kinetics of Elementary
Steps

For the elementary step
A  products
• unimolecular step

v  k A 
For elementary steps involving more than
one reactant
A  B  products v  k AB 
• a bimolecular step
The Kinetics of Elementary
Steps

For the elementary step
A  products
• unimolecular step

v  k A 
For elementary steps involving more than
one reactant
A  B  products v  k AB 
• a bimolecular step

For the step
2 A  B  products
v  k A  B 
2
• a termolecular (three molecule) step.
• Termolecular (and higher) steps are
not that common in reaction
mechanisms.
The Steady-State Approximation

Examine the following simple reaction
mechanism
A
k1
k 1
k2
B P
v p  k 2 B 
Rate of product formation, vp, is proportional
to the concentration of an intermediate.
Applying the Steady State
Approximation (SSA)

Look for the intermediate in the mechanism.
• Step 1 – B is produced.
• Reverse of Step 1 – B is consumed.
• Step 2 – B is consumed.
d B 
 k1 A   k 1 B   k 2 B 
dt
The SSA (Cont’d)

The SSA applied to the intermediate B.
d B 
0
dt
\ k 1 A   k 1 B   k 2 B 
k 1 A 
B  
k 1  k 2
SSA – The Final Step

Substitute the expression for the
concentration of B into the rate law vp.
d P 
 k 2 B 
dt
d P 
k 2 k1
A 
\

dt
k 1  k 2
Competing Reactions

Imagine a reaction with a competing side
reaction.
A B
A C
v b  k a  A 
vc  k b  A 
The Reaction “Yield”

We can calculate the amount of material
produced from the competing reactions.
kJ
 
 kn
n

kJ = the rate constant for the reaction J.
Activated Complex Theory

Consider the following bimolecular
reaction
k
A BC

Presume that the reaction proceeds through the
transition state?
A B
k1
k 1
AB
k
P
The Activated Complex


The species temporarily formed by the
reactant molecules – the activated
complex.
A small fraction of molecules usually
have the required kinetic energy to get to
the transition state
• The concentration of the activated complex is
extremely small.
Transition State Theory (TST)

TST pictures the bi-molecular reaction
proceeding through the activated
complex in a rapid-pre-equilibrium.
A B
k1
k 1
AB
k
P
d P 
 k  AB 
dt
TST (II)

From the thermodynamic equilibrium
constant for the formation of AB↕
p AB
K 
p A pB
p A  RT  A  ; pB  RT B 
p AB  RT  AB 
RT
 AB  
K  A  B 
P
TST (IV)

Assume that we can obtain the Gibbs
energy of activation from K↕.
G  RT ln K
k BT RT G
k2 
 o e
h
P
RT
The Eyring Equation!
TST (V)

Hence, substituting for the Gibbs energy
of activation in terms of the enthalpy and
entropy of activation.
G
 H  T  S
k BT RT H
k2 
 o e
h
P
RT
e
S
R
TST (VI)

Expressing the enthalpy of activation in
terms of the activation energy.
H  E a  2RT
H  E a  RT
bimolcular gas phase reaction
unimolecular gas phase
or solution reaction
TST (VII)

The final result for a bimolecular gas
phase reaction.
k BT RT E a RT S R 2
k2 
 o e
e
e
h
P
E a
E a
S
k
T
RT

2
k 2  Ae RT  B  o e R e RT
h
P
k BT RT S R 2
A
 o e
h
P
TST (VIII)

The final result for a unimolecular gas
phase reaction (or any solution reaction).
k BT RT E a RT S R
kr 
 o e
e
e
h
P
E a
E a
S
k
T
RT

1
k r  Ae RT  B  o e R e RT
h
P
k BT RT S R 1
A
 o e
h
P