3. Chemical Kinetics
 Studies the rates and mechanisms of
chemical reactions.
1
3.1. The Rates of Chemical Reactions
Rate of a reaction:- Change in the concentration/ pressure of reacting
species per unit time
A+BC+D
Rate of change of concentration of the reactant 𝐴 =
 It is a negative quantity
Rate of consumption of reactant 𝐴 = −
 It is a positive quantity
𝑑[𝐴]
𝑑𝑡
𝑑[𝐴]
𝑑𝑡
 Rate of change of concentration of the product 𝐷 =
 It is a positive quantity
Rate of formation of the product 𝐷 =
𝑑[𝐷]
𝑑𝑡
𝑑[𝐷]
𝑑𝑡
 It is a positive quantity
 Rate of change of concentration of reacting species depends on
the coefficient of the species in the balanced chemical equation
2
Example: N2 + 3H2  2NH3
 The rate of consumption of H2 is three times more than the
consumption of N2 .
 The rate of formation of NH3 is two times more than the rate of
consumption of N2.
 For the entire reaction there is only one rate which is a positive
quantity.
 For hypothetical reaction: aA + bB  cC + dD
1 𝑑[𝐴]
1 𝑑[𝐵] 1 𝑑[𝐶] 1 𝑑[𝐷]
𝑟𝑎𝑡𝑒 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 =  = −
=−
=
=
𝑎 𝑑𝑡
𝑏 𝑑𝑡
𝑐 𝑑𝑡
𝑑 𝑑𝑡
Unit of  = moldm-3s-1
Example: If rate of formation of NH3 = 0.16 mMs-1
 rate of consumption of N2 = 0.08 mMs-1
 rate of consumption of H2 = 0.24 mMs-1
 The rate of the reaction = 0.08 mMs-1
3
3.2. Reaction Rate laws
 Reactions may be completed in the time span of picosecond or
less, in microseconds, seconds, minutes, hours, days or years.
 Reaction rates depend on:
 the concentration of reactants
 temperature
catalysts
 the kind of solvent, viscosity and ionic strength of the solution
 intensity of radiation
surface area (for surface reactions)
 Chemical reactions are generally very sensitive to temperature
 Reaction rates must be studied at constant temperature.
 In any kinetic study, the reactant concentration remaining at
various times is the fundamental quantity which requires
measurement
4
3.2.1. Differential Rate law: It is an equation that expresses the rate
of reaction in terms of the concentrations of all the species present in
the overall chemical equation
𝑅𝑎𝑡𝑒 =  = 𝑓( 𝐴 , 𝐵 , … )
 For homogeneous gas-phase reactions, it is often more convenient
to express the rate law in terms of partial pressures,
 = 𝑓(𝑃𝐴, 𝑃𝐵, … )
 Rate law is determined experimentally.
For the reaction: aA + bB  P,
𝟏𝒅 𝑨
𝟏𝒅 𝑩
𝒅[𝑷]
𝒓𝒂𝒕𝒆 = −
=−
=
= 𝒌r[𝑨] [𝑩]
𝒂 𝒅𝒕
𝒃 𝒅𝒕
𝒅𝒕
kr is called rate constant
5
The rate constant
 is a positive value
 is independent of concentrations
 depends on temperature
  and  are order of the reactions with respect to A and B,
respectively.
 The order of a reaction shows the effect of the concentration of
each species on the rate of the chemical reaction.
The values may be zero, positive or negative including fractions
Example: For a first order reaction A  P,  = 𝑘𝑟[𝐴]
If [A] increases by a factor of 2, the rate also increases by a factor of
2
 Overall order of a reaction: The sum of individual orders.
Example: A reaction having a rate law;  = 𝑘r[𝐴]½ [𝐵]
 is ½ order in A, 1st order in B, and the overall order is 3/2 .
6
Some reactions may have indefinite order.
E.g. For a reaction: H2(g) + Br2(g)  2 HBr(g),
3 2
𝑘a 𝐻2 [𝐵𝑟2]
=
𝐵𝑟2 + 𝑘b[𝐻𝐵𝑟]
3.2.2. Integrated rate laws
 An integrated rate law shows how the concentration of a component in a
reaction changes with time.
a) Zero-order reaction
 The rate of a zero-order reaction is independent of the concentration
of the reactant as long as there is a reactant.
Consider: A  P
=−
𝑑𝐴
𝑑𝑡
=𝑘 𝐴
𝑑[𝐴] = −𝑘𝑑𝑡
𝑨 = 𝑨 o − 𝒌𝒕
o
[𝐴]
𝑑[𝐴]
𝐴o
𝑡
= −𝑘 0 𝑑𝑡
Integrated rate law for a zero-order reaction
t = the time the reaction has been in progress, [A]o = initial
concentration of A at t = 0, [A] = concentration of A at any time t
7
A plot of [A] against t gives a straight line with a slope of –k.
Unit of k = moldm-3time-1
Half-life time (t½): The time taken for the concentration of a reactant
to fall to half of its initial value.
For a zero order reaction,
𝐴o
2
= 𝐴 o − 𝑘𝑡½
𝒕½ =
𝑨o
𝟐𝒌
 t1/2 depends on the initial concentration of the reactant.
Note: For a reaction: aA  P , k = akr in all the integrated rate laws
b) First-order reaction
Consider: A  P
𝑑[𝐴]
[𝐴]
𝑑[𝐴]
= 𝑘[𝐴]
𝑑𝑡
[𝐴] 𝑑[𝐴]
𝑡
=
−𝑘
𝑑𝑡
𝐴 o [𝐴]
0
=−
= −𝑘𝑑𝑡
𝒍𝒏 𝑨 = 𝒍𝒏 𝑨 o − 𝒌𝒕
𝒍𝒏
𝑨
𝑨𝒐
= −𝒌𝒕
Integrated rate law for a first – order reaction
𝑨 = 𝑨 o𝒆;𝒌𝒕
8
 A plot of ln[A] against t gives a straight line with a slope of –k.
Unit of k = time-1
 The integrated rate law can be expressed in terms of the product P.
Assuming [P]o = 0, [P] = [A]o –[A], [A] = [A]o - [P]
𝒍𝒏
𝑨 o;[𝑷]
𝑨o
𝑷 = 𝑨 o(𝟏 − 𝒆;𝒌𝒕 )
= −𝒌𝒕
For a first order reaction,
𝒍𝒏𝟐 𝟎. 𝟔𝟗𝟑
𝒕½ =
=
𝒌
𝒌
 t1/2 is independent of initial concentration.
 t1/2 is inversely proportional to the rate constant
Time constant (): The time required for the concentration of a
reactant to fall to 1/e of its initial value.
For a first-order reaction,
𝑘 = −𝑙𝑛
𝐴o
𝑒
𝐴o
=
1
−𝑙𝑛
𝑒
=1
=
𝟏
𝒌
9
Example:
The rate constant for the first-order decomposition of N2O5 in the
reaction 2 N2O5(g)  4 NO2(g) + O2(g) is kr = 3.38 × 10−5 s−1 at 25 °C.
What is the half-life of N2O5?
If the initial partial pressure of N2O5 is 500 Torr, what will its partial
pressure be (i) 50 s, (ii) 20 min after initiation of the reaction?
Solution
0.693
0.693
3𝑠
𝑡½ =
=
=
10.25
×
10
𝑘
2 × 3.38 × 10−5s−1
𝑙𝑛𝑃 = 𝑙𝑛 𝑃o − 𝑘𝑡
i. 𝑙𝑛𝑃 = 𝑙𝑛 500 − 2  3.38 × 10−5𝑠−1 × 50𝑠 = 6.2112
P = 498.3 Torr
ii. 𝑙𝑛𝑃 = 𝑙𝑛 500 − 2  3.38 × 10−5𝑠−1 × 1200𝑠 = 6.1335
P = 461 Torr
10
c) Second-order reaction:
i. Consider A  P, be a 2nd order reaction.
𝑑[𝐴]
= −𝑘 𝐴
𝑑𝑡
2
𝑑[𝐴]
= −𝑘𝑑𝑡
[𝐴]2
𝟏
𝟏
=
+ 𝒌𝒕
[𝑨]
𝑨o
𝟏
𝒕½ =
𝒌𝑨
[𝐴] 𝑑[𝐴]
𝐴 o [𝐴]2
= −𝑘
[𝑨] =
𝑡
𝑑𝑡
0
𝑨o
𝟏:𝒌 𝑨 o𝒕
o
 t½ depends on the initial concentration of the reactant
Unit of k: dm3mol-1time-1
ii. Consider A + B  P, be a second order reaction.
 First order in each of the two reactants, A and B
𝑑[𝐴]
= −𝑘[𝐴][𝐵]
𝑑𝑡
11
𝑑𝐴
𝑑[𝐵]
=−
=−
𝑑𝑡
𝑑𝑡
∆𝐴 = ∆𝐵,
𝐴 𝑜 − 𝐴 = 𝐵 𝑜 − B , 𝐵 = 𝐵 𝑜 − 𝐴 o + [𝐴]
𝑑[𝐴]
= −𝑘[𝐴] 𝐵 o − 𝐴 o + [𝐴]
𝑑𝑡
With [A]o  [B]o
1
[𝐵] 𝐵 o
𝑙𝑛
𝐵 o; 𝐴 o
[𝐴] 𝐴 o
= 𝑘𝑡
𝑙𝑛
[𝐵] 𝐵 o
[𝐴] 𝐴 o
=
𝐵 o− 𝐴
o
𝑘𝑡
3.3. Determination of Rate Laws
a) Graphical Method
For a reaction: A  Products,
 = 𝑘r 𝐴 n
 If the graph of rate vs. [A] gives a straight line with intercept = 0, the
reaction is first order with kr = slope
 If the graph of rate vs. [A]2 gives a straight line with intercept = 0, the
reaction is second order with slope kr
 If the graph of log [A] vs. time gives a straight line with intercept log [A]o,
12
the reaction is first order with k = -slope
b) log/log graphical procedure
For a reaction: A  Products,
 = 𝑘r[𝐴]n
A plot of log rate versus log[A] should be linear with slope n, and
intercept log kr. Hence n and kr can be found.
c) Systematic numerical procedure
 A series of values of the rates at various [reactant] can give the
order and rate constant
For a reaction: A  Products,
 = 𝑘r[𝐴]n
 𝑖𝑛 𝑒𝑥𝑝𝑡. 1 𝑘r 𝑐𝑜𝑛𝑐. 𝑖𝑛 𝑒𝑥𝑝𝑡 1
=
 𝑖𝑛 𝑒𝑥𝑝𝑡. 2 𝑘r 𝑐𝑜𝑛𝑐. 𝑖𝑛 𝑒𝑥𝑝𝑡 2
𝑛
𝑐𝑜𝑛𝑐. 𝑖𝑛 𝑒𝑥𝑝𝑡 1
=
𝑐𝑜𝑛𝑐. 𝑖𝑛 𝑒𝑥𝑝𝑡 2
𝑛
13
d) The initial rate method: Consider a reaction:
A + B  Products
 = 𝑘r 𝐴 [𝐵]
 Change the initial concentration of one reactant, say A, while
keeping the concentration of B constant.
 Measure the initial rates
Order of the reaction with respect to A can be determined
o = keff[A]o
where keff is the effective rate constant = kr[B]o
𝒍𝒐𝒈o = 𝒍𝒐𝒈𝒌eff + 𝐥𝐨𝐠 𝑨 o
 Repeat the procedure by changing the initial concentration of B
while keeping the concentration of A constant
 Order with respect to B can be determined
14
e) Half-life method:
It is applicable if  = 𝑘r[𝐴]𝑛
i. For n = 1, t½ is independent of [A]o.
ii. For n  1
𝑙𝑜𝑔𝑡½ =
2𝑛−1 ;1
𝑙𝑜𝑔
−
𝑛;1 𝑘
𝑛 − 1 log 𝐴
o
 The order of the reaction, n, is determined from the slope.
f) The Isolation Method: If the reaction rate depends on more than
one reactant, keep all except one at high concentration.
 The rate depends on the reactant of the lowest concentration.
For a reaction: A + B  Products
Rate = kr[A][B]
 If [B] >> [A], [B] and hence [B] remains constant during the
reaction,
𝑘𝑟 𝐵  = 𝑘eff = 𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑟𝑎𝑡𝑒 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
15
 Accordingly,  = keff[A]
 The effective rate law is classified as a pseudo-order rate law
Example: If  is 1, the reaction is said to be pseudo-first order
reaction, if 2, pseudo-second order reaction
 The order of the reaction with respect to A can be determined by
determining the rate at different [A].
 From the plot of log  vs. log[A] ,  and keff can be determined
 The order with respect to B can be determined by repeating the
experiment with B still in excess, but at various different values
 The value of keff can be found for each constant excess [B]
𝑘eff = 𝑘r[𝐵]
𝑙𝑜𝑔𝑘eff = 𝑙𝑜𝑔𝑘𝑟 + βlog[𝐵]
A plot of logk eff vs. log[B] should be linear with slope  and intercept
log kr.
Hence, the rate law for the reaction is determined
16
Example: The rate of reaction A  B depends on the concentrations
of both A and B. In one experiment the following data were collected
when the initial concentrations were
[A] = 8  10-4 mol dm-3 , [B] = 0.1 mol dm-3
Rate/ moldm-3min-1 8
4.5 2
0.5
[A]/10-4moldm-3
6
2
8
4
Find the order with respect to A, and the pseudo-rate constant keff. In
a further series of experiments, with B still in large excess, the pseudorate constant keff varied with the concentration of B as follows.
[B]/moldm-3
0.2
keff/ 106mol-1dm3min-1
25.0 37.5 50.0 75.0
0.3
0.4
0.6
17
Find the order of the reaction with respect to B, and hence the overall
order and the true rate constant, kr.
Answer
Dependence of rate on [A]:
 Concentration up by a factor of two, rate up by a factor of four, i.e.
22
 concentration up by a factor of three, rate up by a factor of nine,
i.e. 32
Therefore, reaction is second order in [A]
𝑅𝑎𝑡𝑒 = 𝑘r 𝐵  𝐴 2 = 𝑘eff 𝐴 2
 𝑘eff =
𝑟𝑎𝑡𝑒
𝐴2
𝑘eff =
8.0 𝑚𝑜𝑙𝑑𝑚−3𝑚𝑖𝑛−1
8.010−4 2 𝑚𝑜𝑙2𝑑𝑚−6
= 12.5106𝑚𝑜𝑙−1𝑑𝑚3𝑚𝑖𝑛-1
Using all the data gives an average value of keff = 12.5  106 mol-1 dm3
min-1.
18
𝑘𝑒𝑓𝑓 = 𝑘r 𝐵

𝑙𝑜𝑔𝑘eff = 𝑙𝑜𝑔𝑘r + βlog[𝐵]
Log[B]/moldm-3
-0.699 -0.523 -0.398 -0.222
Logkeff/106mol-1dm3min-1
7.398
7.574
7.699
7.875
B
Linear Fit of Data1_B
7.9
Y = A + Bx
A = 8.097
B=1
7.8
LogK eff
7.7
7.6
7.5
7.4
-0.7
-0.6
-0.5
-0.4
-0.3
-0.2
Log[B]
From the plot,  = 1, kr = 1.25  108mol-2dm6min-1
𝑅𝑎𝑡𝑒 = 𝑘𝑟 𝐴 2 𝐵 where kr = 1.25  108mol-2dm6min-1
19
3.4. Kinetic Analysis for Complex Reactions
3.4.1. Reversible/Opposing reactions
Consider a first order reversible reaction:
A
For the
For the
k1
k-1
B
𝑑𝐴
forward reaction,
𝑑𝑡
𝑑𝐴
reverse reaction,
𝑑𝑡
= −𝑘1[𝐴]
= 𝑘−1[𝐵]
 [A] is reduced by the forward reaction (at a rate k1[A]) and
increased by the reverse reaction (at a rate k-1 [B]).
The net rate of change at any stage of the reaction is:
𝑑[𝐴]
= −𝑘1 𝐴 + 𝑘−1 𝐵
𝑑𝑡
If the initial concentration of A is [A]o, and no B is present initially, then
at all times [A] + [B] = [A]o.
20
Therefore,
𝑑[𝐴]
= −𝑘1 𝐴 + 𝑘−1([A]o−[A]) =−(k1+k−1)[A]+k−1[A]o
𝑑𝑡
Solution:
𝑘−1 + 𝑘1𝑒 ; 𝑘1:𝑘−1 𝑡
[𝐴] =
𝐴o
𝑘1 + 𝑘−1
As t  , the exponential term decreases to zero and the
concentrations reach their equilibrium values
𝐴
𝑒𝑞
𝑘−1 𝐴 𝑜
=
𝑘1 + 𝑘−1
𝑘1 𝐴 o
𝐵 eq = 𝐴 o − 𝐴 eq =
𝑘1 + 𝑘−1
The equilibrium constant is therefore
𝐵
𝐾=
𝐴
𝑘1
=
𝑘−1
eq
eq
At equilibrium, 𝑘1 𝐴
eq
= 𝑘−1 𝐵
eq
f = r
21
Determining the rate constants by Relaxation method:
 Allow the reaction to reach equilibrium and then perturb (e.g.
temperature jump).
 The rate constants change from their initial values
 The rate of restoration to equilibrium gives information on the
rate constants.
Consider:
k1
B
A
k-1
 First order for the forward and reverse reactions
Let the deviation of [A] from its new equilibrium value be x
Therefore
[A] = [A]eq + x and [B] = [B]eq − x
At the new temperature the concentration of A changes as;
𝑑[𝐴]
= −𝑘1 𝐴 + 𝑘−1 B = −𝑘1( 𝐴
𝑑𝑡
eq
+ 𝑥) + 𝑘−1([𝐵]eq − 𝑥)
22
𝑑[𝐴]
= −𝑘1 𝐴 eq + 𝑘−1 𝐵 eq − 𝑘1 + 𝑘−1 𝑥 = − 𝑘1 + 𝑘−1 𝑥
𝑑𝑡
(At equilibrium, k1[A]eq = k-1[B]eq)
d[A]/dt = dx/dt
Therefore,
𝑑𝑥
= − 𝑘1 + 𝑘−1 𝑥
𝑑𝑡
𝑑𝑥
= − 𝑘1 + 𝑘−1 𝑑𝑡
𝑥
𝑥
𝑥o
𝑑𝑥
= − 𝑘1 + 𝑘−1
𝑥
𝑡
𝑑𝑡
0
Solution
𝑡
;
1
𝑘1:𝑘−1
𝑥 = 𝑥o𝑒
where  =
xo is initial departure from equilibrium
 is the relaxation time – the time required to restore equilibrium
23
The rate of restoration to equilibrium is given by
𝒙
𝒙o
𝒕
=𝒆 
;
 From the value of  measured at different x values, the sum, k1 +
k-1 can be known.
 Individual rate constants are determined from the
thermodynamic equilibrium constant, K, where
𝐾=
𝑘1
𝑘−1
 K can be calculated from equilibrium concentrations.
24
3.4.2. Reaction Mechanisms
 Many chemical reactions occur as a sequence of simpler steps,
called elementary reactions with corresponding rate laws that can
be combined into an overall rate law by applying a variety of
approximations.
a) Elementary Reactions
 An elementary reaction involves only a small number of molecules
or ions.
E.g. In the reaction: H + Br2  HBr + Br,
One hydrogen atom attacks one bromine molecule producing one HBr
molecule and one Br atom
 The molecularity of an elementary reaction is the number of
molecules coming together to react in an elementary reaction.
Unimolecular reaction: A single molecule is involved in the
elementary reaction
25
 It is a first order reaction
AP
𝑑[𝐴]
𝑑𝑡
= −𝑘[𝐴]
Bimolecular reaction: Two molecules are involved in the elementary
reaction
 It is a second order reaction
A+BP
𝑑[𝐴]
𝑑𝑡
= −𝑘[𝐴][𝐵]
Note: If the reaction is an elementary bimolecular process, then it has
second-order kinetics, but if the kinetics is second order, then the
reaction might be complex
b) Consecutive/Sequential reactions
 Some reactions proceed through the formation of an intermediate
(I), as in the consecutive uni-molecular reactions
26
k
A
𝑑[𝐴]
𝑑𝑡
a
I
= −𝑘a[𝐴]
k
b
P
𝑑[𝐼]
𝑑𝑡
= 𝑘a 𝐴 − 𝑘b[𝐼]
𝑑[𝑃]
𝑑𝑡
= 𝑘b[𝐼]
Let [I]o = 0, [P]o = 0, [A] + [I] + [P] = [A]0
Solutions
𝐴 = 𝐴 o𝑒 ;𝑘a𝑡
𝑘a
𝐼 =
𝑒 ;𝑘a𝑡 − 𝑒 ;𝑘b𝑡 𝐴
𝑘b − 𝑘a
𝑃 = 𝐴
o
1+
o
𝑘a𝑒 −𝑘b𝑡 ;𝑘b𝑒 −𝑘a𝑡
𝑘b;𝑘a
 The concentration of the intermediate I rises to a maximum and
then falls to zero.
 The concentration of the product P rises from zero towards [A]o,
when all A has been converted to P.
27
Steady-state approximation
At steady-state:
𝑑[𝐼]
=0
𝑑𝑡
𝑘a[𝐴] = 𝑘b[𝐼]
𝑘a
𝑘a
𝐼 =
𝐴 =
𝐴 o𝑒 ;𝑘a𝑡
𝑘b
𝑘b
𝑑[𝑃]
𝑑𝑡
= 𝑘b 𝐼 = 𝑘a 𝐴 = 𝑘a 𝐴 o𝑒 ;𝑘a𝑡
𝑃 = 𝐴
o
1 − 𝑒 ;𝑘a𝑡
Exercise
a) Show that [I] is maximum at a time tmax which is given by
𝑘b
𝑘a
𝑡max =
𝑘b − 𝑘a
𝑙𝑛
b) What is [I]max?
Answer:
𝐼
max
= 𝐴
𝑘b
o 𝑘 ;𝑘
b a
28
c) Pre-equilibrium
 The intermediate is in equilibrium with the reactants
 It can arise when the rate of decay of the intermediate back into
reactants is much faster than the rate at which it forms products;
thus, k-1>> k2.
A + B
k1
I
k2
P
k -1
When A, B, and I are in equilibrium,
𝐾=
𝑑[𝑃]
𝑑𝑡
[𝐼]
𝐴 [𝐵]
𝐾=
𝑘1
𝑘−1
= 𝑘2 𝐼 = 𝑘2𝐾 𝐴 𝐵 =
𝑘1𝑘2
𝑘−1
𝐴 [𝐵]
 In this pre-equilibrium mechanism the final step, I  P, is ratedetermining.
 The preceding steps control the steady concentration of the
intermediate.
29
:
d) Parallel/Side reactions
k1
A
a)
𝑑[𝐴]
𝑑𝑡
B
k2
C
= −𝑘1[𝐴] − 𝑘2[𝐴] = − 𝑘1 + 𝑘2 [𝐴]
Solution: 𝐴 = 𝐴 o𝑒 ;
b)
𝑑[𝐵]
𝑑𝑡
= 𝑘1[𝐴] = 𝑘1 𝐴 o𝑒 ; 𝑘1:𝑘2
𝑘1 𝐴 o
𝑘1:𝑘2
Solution: [𝐵] =
c)
𝑑[𝐶]
𝑑𝑡
𝑘1:𝑘2 𝑡
1 − 𝑒;
= 𝑘2[𝐴] = 𝑘2 𝐴 o𝑒 ; 𝑘1:𝑘2
Solution: [C] =
From [B] and [C],
[𝑩]
[𝑪]
=
𝑘2 𝐴 o
𝑘1:𝑘2
𝒌1
𝒌2
𝑡
𝑘1:𝑘2 𝑡
𝑡
1 − 𝑒;
𝑘1:𝑘2 𝑡
30
e) Chain reactions
 Reactants are converted into reactive intermediates (usually
radicals).
 The intermediates further react to give products.
Steps:
i. Initiation: Free radicals are formed from ordinary molecules.
ii. Propagation: The free radicals attack other molecules producing
free radicals and products.
iii. Inhibition: The free radicals attack the products thereby
decreasing the rate.
iv. Termination: Free radicals combine to give normal molecules,
thereby stopping the reaction.
31
Consider the reaction:
CH3CHO (g)  CH4 (g) + CO (g)  = 𝑘[CH3CHO]3/2
experimentally determined
Proposed reaction mechanism
 Test the proposed mechanism if it leads to the observed rate law.
Steady-state approximation:
 The rate of change of the intermediates is zero.
32
Solution:
At steady- state
Rate of formation of CH4 is therefore,
Where
½
𝑘i
𝑘 = 𝑘p
2𝑘t
 It is in agreement with the three-halves order observed
experimentally.
33
f) Catalyzed Reactions
i) Acid Base Catalysis
 The mechanism for acid catalysis may be illustrated by the
following reaction
k1
+
a) AH + S
b) SH
+
k-1
k2
SH+ + A
products
Reaction b is of two types
+
i. SH + H2O
+
ii. SH + A
k2
k2
H3O+ + product
AH+ + Product
Case i.
𝑑𝑃
𝑑𝑡
= 𝑘2[𝑆𝐻+]
34
Steady state approximation:
𝑑[𝑆𝐻+]
𝑑𝑡
[𝑆𝐻+]
𝑑𝑃
𝑑𝑡
= 0 = 𝑘1 𝑆 𝐴𝐻+ − 𝑘−1 𝑆𝐻+ 𝐴 − 𝑘2[𝑆𝐻+]
=
= 𝑘2
𝑘1[𝑆][𝐴𝐻+]
𝑘−1 𝐴 :𝑘2
𝑆𝐻+
=
𝑘1𝑘2 𝑆 [𝐴𝐻+]
𝑘−1 𝐴 :𝑘2
Cases:
i. If k2 >> k-1[A]
𝑑𝑃
= 𝑘1 𝑆 [𝐴𝐻+]
𝑑𝑡
Therefore, the rate depends on the concentration of the catalyst
35
ii. If k2 << k-1[A]
𝑑𝑃
𝑑𝑡
= 𝑘2 𝑆𝐻+ =
𝑘1𝑘2 𝑆 [𝐴𝐻+]
𝑘−1[𝐴]
For the acid dissociation,
AH+
𝐾=
A + H+
𝐴 [𝐻+]
[𝐴𝐻+]
𝐴𝐻+ =
𝐴 [𝐻+]
𝐾
𝑑𝑃 𝑘1𝑘2 𝑆 [𝐻+]
=
𝑑𝑡
𝑘−1𝐾
 The rate specifically depends on the concentration of H+
Exercise
Determine the rate law considering case bii.
36
ii) Enzyme Catalysis
Michalis – Menten Equation
 Consider the following mechanism
k1
a) E + S
ES
k-1
b) ES
Let;
𝑟𝑎𝑡𝑒 =
At steady state,
𝑑[𝐸𝑆]
𝑑𝑡
k2
E + product
= 𝑘2[𝐸𝑆]
=0
𝑘1 𝐸 𝑆 = 𝑘−1 𝐸𝑆 + 𝑘2[𝐸𝑆]
𝐸 o = 𝐸 + [𝐸𝑆]
𝑘1 𝐸 o 𝑆
[𝐸𝑆] =
𝑘−1 + 𝑘2 + 𝑘1[𝑆]
37
[𝐸𝑆] = 𝑘
𝐸 o[𝑆]
−1+𝑘2:[𝑆]
𝑘1
Where
𝐾m =
Hence,
𝑘−1:𝑘2
𝑘1
[𝐸𝑆] =
𝐸 o[𝑆]
𝐾𝑚:[𝑆]
= Michalis constant
𝒌2 𝑬 o[𝑺]
𝒌2 𝑬 o
𝒓𝒂𝒕𝒆 =
=
𝑲𝒎 + [𝑺] 𝟏 + 𝑲m
[𝑺]
Cases
1. If [S] >> Km
Rate = k2[E]o = max
Rate becomes independent of [S]
2. If [S] << Km
𝒌𝟐
𝑹𝒂𝒕𝒆 = =
𝑬 o[𝑺]
𝑲m
Other form:
𝟏

=
𝟏
max
+
𝑲m
max
𝟏
𝑺o
38
3.5. The Arrhenius Equation
 Chemical reactions usually go faster as the temperature is raised.
 An increase in temperature results with an increase in reaction
rate constant
𝑬
𝒌r = 𝑨𝒆
;𝑹𝑻a
Arrhenius equation
A = frequency factor (pre-exponential factor), Ea = Activation energy
Ea and A are called Arrhenius parameters
 Experimentally for many reactions a plot of ln kr against 1/T gives
a straight line with a negative slope
𝐸𝑎
𝑙𝑛𝑘𝑟 = 𝑙𝑛𝐴 −
𝑅𝑇
From y-intercept, A can be determined and from the slope (-Ea/R),
Ea can be obtained.
39
Note:
A high activation energy signifies that the rate constant depends strongly on
temperature.
 If a reaction has zero activation energy, its rate is independent of
temperature.
 A negative activation energy indicates that the rate decreases as the
temperature is raised.
Example: The activation energy for a reaction is 50kJmol-1. By how much will
the reaction rate constant increase when the reaction temperature
changes from 25 to 36 oC?
Solution:
𝐸a 1
1
𝑘
𝐸
1
−
, 𝑙𝑛 r2 = a
𝑅 𝑇1
𝑇2
𝑘r1
𝑅 𝑇1
50×105 1
1
−
= 0.7184
8.314 298
309
𝑙𝑛𝑘r2 − 𝑙𝑛𝑘r1 =
𝑘r2
𝑙𝑛
=
𝑘𝑟1
𝑘r2
=2,
𝑘r1
−
1
𝑇2
Kr2 = 2kr1
 The rate constant has increased by a factor of 2.
40
3.6. Theories of reaction rate
3.6.1. Collision Theory
 The rate of a chemical reaction is equal to the number of headon collisions per unit volume per unit time multiplied by the
fraction of all collisions that have the required energy to yield
products.
Consider the bimolecular elementary reaction:
A + B P
 = 𝑘2 𝐴 [𝐵]
 The rate is proportional to:
 the rate of collision and therefore to the mean speed of the
molecules,
their collision cross-section,, and
the number densities of A and B (NA[A], NA[B]).
41
Collision density (ZAB): Number of collisions per unit volume per
unit time.
½
8𝑘𝑇
𝑍AB = 
𝑁A2 𝐴 [𝐵]
𝜋𝑀
Where  = collision-cross section = d2 (d = effective molecular
diameter) ((AB) = (rA+rB)2}
 All collisions within  do not lead to products.
 Only those within reaction cross-section (*) lead to products.
* = P where P is a steric factor.
Hence,
𝑍AB = 𝑃
8𝑘𝑇 ½
𝑁A2
𝜋𝑀
𝐴 [𝐵]
 Let f be fraction of collisions that satisfy the energy requirement.
This makes the rate to be proportional to the Boltzmann factor,
e-Ea/RT
42
𝑑[𝐴]
−
𝑑𝑡
=
𝑍AB
𝑓
𝑁A
 Dividing by the Avogadro’s number converts the number of
molecules to molar concentrations.
Hence,
 = 𝑃
𝐸a
8𝑘𝑇 ½
−
𝑁Ae 𝑅𝑇
𝜋𝑀
𝐴 [𝐵]
Therefore,
k2 =
𝐸a
8𝑘𝑇 ½
−
𝑃
𝑁Ae 𝑅𝑇
𝜋𝑀
8𝑘𝑇 ½
Let 𝐴 = 𝜎
𝜋𝑀
𝒌2 = 𝑷𝑨𝒆
𝑁𝐴 - Arrhenius parameter
𝑬
;𝑹𝑻a
43
3.6.2. Transition State theory
 The rate of a reaction is equal to the rate with which reactants
pass over the energy barrier multiplied by the concentration of the
activated complex.
Consider the elementary reaction:
C*
mA
P
C* is the activated complex.
𝑑𝐴
=−
= 𝑘2 𝐴
𝑑𝑡
∗
𝐾C =
𝑚
= 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑝𝑎𝑠𝑠𝑎𝑔𝑒 𝑜𝑣𝑒𝑟 𝑒𝑛𝑒𝑟𝑔𝑦 𝑏𝑎𝑟𝑟𝑖𝑒𝑟 × 𝐶∗ = 𝑘∗[𝐶∗]
[𝐶∗]
𝐴m
Where 𝐾𝑐 ∗ is the equilibrium constant for the formation of the
complex.
44
 The rate of passage of the complex through the energy barrier is
proportional to the vibrational frequency of the complex along the
reaction coordinate.
𝑘∗ = 
 = vibrational frequency, k* = rate constant of the complex
decomposition,  = transmission coefficient (~1)
=
𝑘𝑇
ℎ
where h = Planck constant, k = Boltzmann constant
Therefore,
𝑘𝑇
∗
𝑘 =
𝑕
Hence,
𝑘𝑇
=
𝐶∗ = 𝑘2 𝐴
ℎ
Therefore,
𝑘𝑇𝐾C∗
k2 =
ℎ
m
=
𝑘𝑇𝐾C∗
ℎ
𝐴
m
Eyring equation
45