Kinetika Kimia
(Chemical Kinetics)
Jaslin Ikhsan, Ph.D.
Kimia FMIPA UNY
Course Summary.
• Dependence of rate on concentration.
• Experimental methods in reaction
kinetics.
• Kinetics of multistep reactions.
• Dependence of rate on temperature.
Recommended reading.
1. P.W. Atkins, Kimia Fisika (terjemahan), jilid 2, edisi keempat,
Jakarta: Erlangga, 1999.
2. P.W. Atkins, Physical Chemistry, 5th ed., Oxford:
Oxford University Press, 1994
4. Arthur M. Lesk, Introduction to Physical Chemistry
Edited by Foxit PDF Editor
Copyright (c) by Foxit Software Company, 2004
For Evaluation Only.
3. Gordon M. Barrow, Physical Chemistry
Chemical Kinetics.
Lecture 1.
Review of Basic concepts.
Reaction Rate: The Central Focus of
Chemical Kinetics
Chemical reaction kinetics.
•
•
•
Chemical reactions involve the
forming and breaking of
chemical bonds.
Reactant molecules (H2, I2)
approach one another and
collide and interact with
appropriate energy and
orientation. Bonds are
stretched, broken and formed
and finally product molecules
(HI) move away from one
another.
How can we describe the rate
at which such a chemical
transformation takes place?
reactants
products
H 2 ( g ) + I 2 ( g ) → 2HI ( g )
• Thermodynamics tells us all
about the energetic feasibility
of a reaction : we measure the
Gibbs energy ∆G for the chemical
Reaction.
• Thermodynamics does not tell us
how quickly the reaction will
proceed : it does not provide
kinetic information.
Basic ideas in reaction kinetics.
•
Chemical reaction kinetics deals with the rate of velocity of chemical
reactions.
•
We wish to quantify
•
These objectives are accomplished using experimental measurements.
•
Chemical reactions are said to be activated processes : energy (usually
thermal (heat) energy) must be introduced into the system so that
chemical transformation can occur. Hence chemical reactions occur
more rapidly when the temperature of the system is increased.
•
In simple terms an activation energy barrier must be overcome before
reactants can be transformed into products.
– The velocity at which reactants are transformed to products
– The detailed molecular pathway by which a reaction proceeds (the reaction
mechanism).
•
Reaction Rate.
What do we mean by the term
reaction rate?
–
•
How may reaction rates be
determined ?
–
–
•
•
•
The term rate implies that something
changes with respect to something
else.
The reaction rate is quantified in
terms of the change in concentration
of a reactant or product species with
respect to time.
This requires an experimental
measurement of the manner in which
the concentration changes with time
of reaction. We can monitor either
the concentration change directly, or
monitor changes in some physical
quantity which is directly proportional
to the concentration.
The reactant concentration
decreases with increasing time, and
the product concentration increases
with increasing time.
The rate of a chemical reaction
depends on the concentration of each
of the participating reactant species.
The manner in which the rate
changes in magnitude with changes in
the magnitude of each of the
participating reactants is termed the
reaction order.
Reactant
concentration
Product
concentration
d [R] d [P]
RΣ = −
=
dt
dt
Net reaction rate
Units : mol dm-3 s-1
[P]t
[R]t
time
Rate, rate equation and reaction order : formal
definitions.
• The reaction rate (reaction velocity) R is quantified in terms of
changes in concentration [J] of reactant or product species J
with respect to changes in time. The magnitude of the reaction rate
changes as the reaction proceeds.
∆[J ] 1 d [J ]
RJ =
lim
=
t
∆
→
0
∆t
υJ
υ J dt
1
2 H 2 ( g ) + O2 ( g ) → 2 H 2O( g )
R=−
1 d [H 2 ]
d [O2 ] 1 d [H 2O ]
=−
=
2 dt
dt
2 dt
• Note : Units of rate :- concentration/time , hence RJ has units mol dm-3s-1 .
υJ denotes the stoichiometric coefficient of species J. If J is a reactant υJ
is negative and it will be positive if J is a product species.
• Rate of reaction is often found to be proportional to the molar
concentration of the reactants raised to a simple power (which
need not be integral). This relationship is called the rate equation.
The manner in which the reaction rate changes in magnitude with
changes in the magnitude of the concentration of each participating
reactant species is called the reaction order.
Problem Example:
The rate formation of NO in the reaction:
2NOBr(g) ----> 2NO(g) + Br2(g)
was reported as 1.6 x 10-4 mol/(L s).
What is the rate of reaction and the rate of consumption
of NOBr ?
Log R
Log R
Slope = α
Slope = β
Log [A]
Log [B]
k
xA + yB 
→
Products
empirical rate
equation (obtained
from experiment)
rate constant k
R=−
stoichiometric
coefficients
Rate equation can not in
general be inferred from
the stoichiometric equation
for the reaction.
1 d [ A]
1 d [B ]
α
β
=−
= k [ A] [B ]
x dt
y dt
α, β = reaction
orders for the
reactants (got
experimentally)
Different rate equations
imply different mechanisms.
H 2 + I 2 → 2 HI
d [HI ]
= k [H 2 ][I 2 ]
dt
H 2 + Br2 → 2 HBr
R=
d [HBr ] k [H 2 ][Br2 ]
R=
=
k ′[HBr ]
dt
1+
[Br2 ]
1/ 2
H 2 + Cl2 → 2 HCl
d [HCl ]
1/ 2
= k [H 2 ][Cl2 ]
R=
dt
H 2 + X 2 → 2 HX
X = I , Br , Cl
• The rate law provides an important guide
to reaction mechanism, since any proposed
mechanism must be consistent with the
observed rate law.
• A complex rate equation will imply a complex
multistep reaction mechanism.
• Once we know the rate law and the rate
constant for a reaction, we can predict the
rate of the reaction for any given composition
of the reaction mixture.
• We can also use a rate law to predict the
concentrations of reactants and products at
any time after the start of the reaction.
Zero order kinetics. The reaction proceeds at the same rate R
regardless of concentration.
Rate equation :
da
R=−
=k
dt
a = a0 when t = 0
a (t ) = − kt + a0
0
units of rate constant k :
mol dm-3 s-1
[ A]
integrate
using initial
condition
half life : t = τ 1/ 2 when a =
a
τ 1/ 2 =
slope = -k
a0
2k
a0
2
τ 1 / 2 ∝ a0
τ 1/ 2
a0
diagnostic
plot
R ∝ [ A]
R
slope =
t
a0
1
2k
k
A
→
products
First order kinetics.
Initial
concentration a0
rate = −
First order differential
rate equation.
da
−
= ka
dt
d [ A] t
dt
Initial condition
t = 0 a = a0
Solve differential
equation
Separation
of variables
For a first order reaction the relationship:
(rate) t ∝ [ A] t
(rate) t = k [ A] t
a (t ) = a0 e − kt = a0 exp[− kt ]
is valid generally for any time t.
k is the first order
rate constant, units: s-1
Reactant concentration
as function of time.
First order kinetics.
Half life τ1/2
t = τ 1/ 2
θ = θ1/ 2
τ 1/ 2 =
k → s −1
a (t ) = a0 e − kt = a0 exp[− kt ]
a (t )
u=
= exp[− θ ]
a0
θ = kt
a0
2
u = 1/ 2
a=
ln 2 0.693
=
k
k
Mean lifetime of reactant molecule
1
τ=
a0
∫
∞
0
1
a(t )dt =
a0
∫
∞
0
a0 e − kt dt =
1
k
First order kinetics: half life.
u=
a
a0
u =1
t = τ 1/ 2
In each successive period
of duration τ1/2 the concentration
of a reactant in a first order reaction
decays to half its value at the start
of that period. After n such periods,
the concentration is (1/2)n of its
initial value.
a = a0 / 2
u = 0 .5
u = 0.25
u = 0.125
τ 1/ 2
τ 1/ 2 =
a0
ln 2 0.693
=
k
k
half life independent of
initial reactant concentration
1st order kinetics
u (θ )
a (t ) = a0 e
u (θ ) = e −θ
θ = kt
2nd order kinetics
− kt
u (θ )
a (t ) =
a0
1 + ka0t
u (θ ) =
θ = ka0t
1
1+θ
Second order kinetics: equal
reactant concentrations.
da
= ka 2
dt
t = 0 a = a0
−
1
a
separate variables
integrate
1
1
= kt +
a
a0
half life t = τ 1/ 2 a = a0
2
τ 1/ 2 =
k
2A 
→
P
1
ka0
1
τ 1/ 2 ∝
a0
τ 1/ 2 ↓ as a0 ↑
τ 1/ 2
dm3mol-1s-1
slope = k
t
a0
a (t ) =
1+ ka0t
a0
rate varies as
square of reactant
concentration
1st and 2nd order kinetics : Summary .
Reaction
k1
A →
Products
k2
2 A →
Products
Differential
rate equation
Concentration
variation with
time
da
−
= k1a
dt
a (t ) =
a0 exp[− k1t ]
a (t ) =
da
2
−
= k2 a
a0
dt
Diagnostic
Equation
ln a(t ) = −k1t + ln a0
1
1
= k 2t +
a (t )
a0
Half
Life
τ 1/ 2 =
ln 2
k1
τ 1/ 2 =
1
k 2 a0
1 + k 2 a0t
ln a(t)
Slope = - k1 1/a(t)
τ1/2
Slope = k2
t
t
1st order
2nd order
a0
Diagnostic
Plots .
n th order kinetics: equal reactant
concentrations.
da
−
= ka n
dt
t = 0 a = a0
separate variables
integrate
a
n ≠1
n = 0, 2,3,…..
k
nA 
→
P
1
a n −1
1
n −1
= (n − 1)kt +
1
a0
n −1
rate constant k
obtained from slope
Half life
τ 1/ 2 =
n −1
2 −1
(n − 1)ka0 n−1
τ 1/ 2 ∝ a01− n
n > 1 τ 1/ 2 ↓ as a0 ↑
n < 1 τ 1/ 2 ↑ as a0 ↑
 2 −1 
ln τ 1/ 2 = ln 
 − (n − 1) ln a0
(
)
n
1
k
−


n −1
slope = (n − 1)k
t
ln τ 1/ 2
slope = −(n − 1)
reaction order n determined
from slope
ln a0
Summary of kinetic results.
t = 0 a = a0
k
nA 
→
P
t = τ 1/ 2
a=
Rate equation
Reaction
Order
da
R=−
dt
Integrated
expression
Units of k
mol dm-3s-1
a0
2
Half life
τ1/2
0
k
a(t ) = −kt + a0
1
ka
a 
ln  0  = kt
 a (t ) 
s-1
ln 2
k
2
ka
2
1
1
= kt +
a (t )
a0
dm3mol-1s-1
1
ka0
ka
3
1
1
=
2
kt
+
2
2
a(t )
a0
dm6mol-2s-1
3
2
2ka0
ka
n
1
3
n
a n −1
= (n − 1)kt +
1
a0
n −1
a0
2k
1  2 n −1 − 1 


n − 1  ka 0 n −1 
Problem Examples:
1. A first order reaction is 40% complete at the end of 1 h.
What is the value of the rate constant?
In how long willl the reaction be 80 % complete ?
2. The half-life of the radioactive disintegration of radium is
1590 years. Calculate the decay constant. In how many
years will three-quarters of the radium have undergone
decay ?
3. Derive the rate equation for reaction whose orders are:
(a). one-half, (b). three and a-half, (c). 4, and (d). n !
Chemical Kinetics
Lecture 2.
Kinetics of more complex
reactions.
Jaslin Ikhsan, Ph.D.
Kimia FMIPA
Universitas Negeri Jogjakarta
Tujuan Perkuliahan:
1. Menurunkan Rate Law dari Kinetika
Reaksi yang lebih Kompleks,
2. Menjelaskan Consecutive Reactions:
a. Rate Determining Steps,
b. Steady State Approximation.
Reference:
1. P.W. Atkins, Physical Chemistry, 5th ed,
Oxford: 1994
2. Ira N. Levine, Physical Chemistry
Second order kinetics:
Unequal reactant concentrations.
A+ B 
→ P
k
rate equation
R=−
da
db dp
=−
=
= kab
dt
dt dt
initial conditions
t = 0 a = a0
b = b0
a0 ≠ b0
integrate using
partial fractions
1
F (a, b ) =
b0 − a0
  b b0 
 = kt
ln
  a a0  
dm3mol-1s-1
F (a, b )
slope = k
t
Consecutive Reactions .
•Mother / daughter radioactive
decay.
218
Mass balance requirement:
Po→ 214 Pb→ 214 Bi
k1 = 5 × 10 −3 s −1
k1
k2
A →
X →
P
k 2 = 6 × 10 − 4 s −1
p = a0 − a − x
The solutions to the coupled
equations are :
3 coupled LDE’s define system :
a (t ) = a0 exp[− k1t ]
da
= − k1a
dt
dx
= k1a − k 2 x
dt
dp
= k2 x
dt
x(t ) =
k1a0
{ exp[− k1t ] − exp[− k2t ] }
k 2 − k1
p(t ) = a0 − a0 exp[− k1t ] −
k1a0
{ exp[− k1t ] − exp[− k2t ] }
k 2 − k1
We get different kinetic behaviour depending
on the ratio of the rate constants k1 and k2
Consecutive reaction : Case I.
Intermediate formation fast, intermediate decomposition slow.
Case I .
κ=
k2
<< 1
k1
TS II
k 2 << k1
TS I
I : fast
∆GI‡
<<
II : slow
rds
∆GII‡
energy
k1
k2
A →
X →
P
∆GI‡
∆GII‡
A
X
P
reaction co-ordinate
Step II is rate determining
since it has the highest
activation energy barrier. The reactant species A will be
more reactive than the intermediate X.
Case I .
κ=
k2
<< 1
k1
k 2 << k1
k1
k2
A →
X →
P
a
u=
a0
x
v=
a0
p
w=
a0
Normalised concentration
1.2
κ = k2/ k1 = 0.1
1.0
u = a/a0
v = x/a0
Intermediate X
0.8
w = p/a0
0.6
Reactant A
0.4
Product P
0.2
0.0
0
2
4
6
8
10
τ = k 1t
Initial reactant A more
reactive than intermediate X .
Concentration of intermediate
significant over time course of
reaction.
Consecutive reactions Case II:
Intermediate formation slow, intermediate decomposition fast.
Case
CaseIIII..
κ=
k2
>> 1
k1
k1
k2
A →
X →
P
key parameter
Intermediate X fairly reactive.
[X] will be small at all times.
k 2 >> k1
k1
k2
A →
X →
P
∆GI‡ >>
TS I
II : fast
∆GII‡
energy
I : slow rds
k2
κ=
k1
TS II
∆GI‡
∆GII‡
A
Step I rate determining
since it has the highest
activation energy barrier.
X
P
reaction co-ordinate
1.2
Case
CaseIIII..
κ=
normalised concentration
k1
k2
A →
X →
P
k2
>> 1
k1
k 2 >> k1
Intermediate X
is fairly reactive.
Concentration of
intermediate X
will be small at
all times.
1.0
Product P
0.8
u=a/a0
v =x/a0
0.6
w=p/a0
Reactant A
0.4
0.2
κ = k 2/k1 = 10
Intermediate X
0.0
0
2
4
6
8
10
τ = k1t
normalised concentration
1.2
u=a/a0
1.0
Intermediate concentration
is approximately constant
after initial induction period.
v=x/a0
w=p/a0
0.8
P
A
0.6
0.4
X
0.2
0.0
0.0
0.2
0.4
0.6
τ = k1t
0.8
1.0
Rate Determining Step
k1 >> k 2
A → X → P
k1
Fast
k2
Slow
• Reactant A decays rapidly, concentration of intermediate species X
is high for much of the reaction and product P concentration rises
gradually since X--> P transformation is slow .
k 2 >> k1
Rate Determining
Step
A → X → P
k1
k2
Slow
Fast
• Reactant A decays slowly, concentration of intermediate species X
will be low for the duration of the reaction and to a good approximation
the net rate of change of intermediate concentration with time is zero.
Hence the intermediate will be formed as quickly as it is removed.
This is the quasi steady state approximation (QSSA).
Parallel reaction mechanism.
k1
A →
X
k2
• We consider the kinetic analysis of a concurrent
A →
Y
or parallel reaction scheme which is often met in
k1, k2 = 1st order rate constants
real situations.
• A single reactant species can form two
We can also obtain expressions
distinct products.
for the product concentrations
st
We assume that each reaction exhibits 1 order x(t) and y(t).
kinetics.
• Initial condition : t= 0, a = a0 ; x = 0, y = 0
.• Rate equation:
da
= k1a + k 2 a = (k1 + k 2 )a = kΣ a
R=−
dt
a (t ) = a0 exp[− kΣt ] = a0 exp[− (k1 + k 2 )t ]
• Half life: τ 1/ 2 = ln 2 = ln 2
kΣ
k1 + k 2
dx
= k1a = k1a0 exp[− (k1 + k 2 )t ]
dt
x(t ) = k1a0
x(t ) =
∫ exp[− (k
t
1
0
+ k 2 )t ] dt
k1a0
{1 − exp[− (k1 + k2 )t ]}
k1 + k 2
dy
= k 2 a = k 2 a0 exp[− (k1 + k 2 )t ]
dt
y (t ) = k 2 a0
y (t ) =
∫ exp[− (k
t
0
1
+ k 2 )t ] dt
k 2 a0
{1 − exp[− (k1 + k2 )t ]}
k1 + k 2
x(t ) k1
• All of this is just an extension of simple Final product analysis
=
Lim
st
yields
rate
constant
ratio.
t
→
∞
y
(
t
)
k2
1 order kinetics.
Parallel Mechanism: k1 >> k2
normalised concentration
1.0
Theory
k
κ = 2 = 0.1
k1
k1
v(∞ )
= 10 =
k2
w(∞)
κ = 0.1
a(t)
0.8
v(∞ ) = 0.9079
x(t)
u(τ)
v(τ)
w(τ)
0.6
0.4
κ = k2 /k1
/
0.2
y(t)
w(∞ ) = 0.0908
0.0
0
1
2
3
τ = k1t
4
5
6
Computation
v(∞ ) 0.9079
≅
= 9.9989
w(∞ ) 0.0908
Parallel Mechanism: k2 >> k1
normalised concentration
1.0
Theory
k2
= 10
k1
k1
v(∞ )
= 0.1 =
k2
w(∞)
w(∞) ≅ 0.9091
u(τ)
v(τ)
w(τ)
0.6
0.4
a(t)
0.2
x(t)
v(∞) ≅ 0.0909
0.0
0
κ=
y(t)
κ = 10
0.8
1
2
3
τ = k1t
4
5
6
Computation
v(∞ ) 0.0909
≅
= 0.0999
w(∞ ) 0.9091
Reaching Equilibrium on the
Macroscopic and Molecular Level
NO2
N2O4
N2O4 (g)
colourless
2 NO2 (g)
brown
Chemical Equilibrium :
a kinetic definition.
•
•
Countless experiments with chemical
systems have shown that in a state of
equilibrium, the concentrations of
reactants and products no longer change
with time.
This apparent cessation of activity occurs
because under such conditions, all
reactions are microscopically reversible.
We look at the dinitrogen tetraoxide/
nitrogen oxide equilibrium which
occurs in the gas phase.
N2O4 (g)
colourless
Kinetic analysis.
r
R = k [N 2O4 ]
s
2
R = k ′[NO2 ]
Valid for any time t
2 NO2 (g)
brown
Concentrations vary
with time
[NO2 ] t ↑
[N 2O4 ] t
concentration
•
Equilibrium:
r s
t →∞
R=R
2
k [N 2O4 ]eq = k ′[NO2 ]eq
[NO2 ]eq2 k
= =K
[N 2O4 ] k ′
↓
Kinetic
regime
Concentrations time
invariant
[NO2 ] eq
[N 2O4 ] eq
Equilibrium
state
NO2
t →∞
N2O4
time
Equilibrium
constant
First order reversible reactions :
understanding the approach to chemical equilibrium.
Rate equation
t = 0 a = a0
Rate equation in normalised form
du
1
+u =
dτ
1+θ
b=0
Mass balance condition
Solution produces the concentration expressions
a + b = a0
u (τ ) =
v(τ ) =
Introduce normalised variables.
a
a0
v=
b
a0
k'
B
τ = 0 u =1 v = 0
Initial condition
u=
A
u + v =1
da
= −ka + k ′b
dt
∀t
k
τ = (k + k ′)t θ =
k
k′
1
{1 + θ exp[− τ ]}
1+θ
θ
1+θ
{1 − exp[− τ ]}
Q(τ ) =
Reaction quotient Q
v(τ )
=θ
u (τ )
 1 − exp[− τ ] 


[
]
+
−
1
θ
exp
τ


First order reversible reactions: approach to equilibrium.
Kinetic
regime
1.0
concentration
0.8
Equilibrium
Product B
0.6
K = Q(τ → ∞ )
u (τ)
v (τ)
0.4
Reactant A
=
0.2
0.0
0
2
4
τ = (k+k')t
6
8
v(∞ )
u (∞ )
Understanding the difference between reaction quotient Q and
Equilibrium constant K.
12
Approach to
Equilibrium
Q<θ
10
θ = 10
Q(τ)
8
Equilibrium
Q=K=θ
6
4
2
0
0
2
4
6
8
τ = (k+k')t
Q(τ ) =
v(τ )
=θ
u (τ )
 1 − exp[− τ ] 


1 + θ exp[− τ ]
k
t → ∞ Q → K =θ =
k′
K=
v(∞ )
u (∞ )
H H
H H
C
C
C
H X
C
H X
n
Chemical Kinetics.
Lecture 3/4.
Application of the Steady State
Approximation: Macromolecules
and Enzymes.
Quasi-Steady State Approximation.
• Detailed mathematical analysis of complex
QSSA
reaction mechanisms is difficult.
Some useful methods for solving sets of
coupled linear differential rate equations
include matrix methods and Laplace Transforms.
• In many cases utilisation of the quasi steady
state approximation
induction
leads to a considerable simplification in the
period
kinetic analysis.
k1
k2
A →
X →
P
Consecutive reactions k1 = 0.1 k2 .
The QSSA assumes that after an initial
induction period (during which the
concentration x of intermediates X rise
from zero), and during the major part
of the reaction, the rate of change of
concentrations of all reaction
intermediates are negligibly small.
Mathematically , QSSA implies
dx
= RX formation − RX removal ≅ 0
dt
RX formation = RX removal
Normalised concentration
1.0
u(τ)
A
P
0.5
w(τ)
v(τ)
0.0
0.0
0.5
1.0
τ = k1t
intermediate X
concentration
approx. constant
Consecutive reaction mechanisms.
k1
A
X
k-1
Rate
equations
u=
k2
P
du
= −u + κ v
dτ
dv
= u − (κ + φ ) v
dτ
dw
=φ v
dτ
a
a0
v=
x
a0
w=
p
a0
∀τ u + v + w = 1
τ = 0 u =1 v = w = 0
k
κ = −1
k1
k
φ= 2
k1
τ = k1 t
Definition of normalised variables
and initial condition.
Step I is reversible, step II is
Irreversible.
Coupled LDE’s can be solved via Laplace
Transform or other methods.
1
{(κ + φ − α ) exp[− α τ ] − (κ + φ − β ) exp[− β τ ]}
β −α
1
{exp[− α τ ] − exp[− β τ ]}
v(τ ) =
β −α
1
{β exp[− α τ ] − α exp[− β τ ]}
w(τ ) = 1 −
β −α
u (τ ) =
Note that α and β are composite quantities containing
the individual rate constants.
αβ = φ
α + β = 1+ κ + φ
QSSA assumes that
dv
≅0
dτ
u ss − (κ + φ )vss = 0
u
vss ≅ ss
κ +φ
du ss
φ
=−
u ss
dτ
κ +φ

φ
u ss ≅ exp −
 κ +φ

τ

  φ  
 τ 
exp − 
 κ +φ  
vss ≅
κ +φ
Using the QSSA we can develop more
simple rate equations which may be
integrated to produce approximate
expressions for the pertinent concentration
profiles as a function of time.
The QSSA will only hold provided that:
• the concentration of intermediate is small
and effectively constant,
and so :
• the net rate of change in intermediate
concentration wrt time can be set equal to
zero.
  φ  
dwss
φ
 τ 
exp − 
≅ φ vss =
κ +φ
κ
φ
dτ
+
 
 
τ
  φ  
 τ  dτ
exp
− 
κ + φ ∫0
κ
φ
+
 
 
  φ  
 τ 
= 1 − exp − 
κ
φ
+
 
 
wss ≅
φ
normalised concentration
1.2
1.0
A
P
A
0.4
X
0.0
0.01
0.1
1
10
k2
P
Concentration versus log time curves
for reactant A, intermediate X and
product P when full set of coupled
rate equations are solved without
any approximation.
k-1 >> k1, k2>>k1 and k-1 = k2 = 50.
The concentration of intermediate X is
very small and approximately constant
throughout the time course of the
experiment.
u(τ)
v(τ)
w(τ)
0.2
X
k-1
0.8
0.6
k1
100
log τ
QSSA will hold when concentration
of intermediate is small and constant.
Hence the rate constants for getting
rid of the intermediate (k-1 and k2)
must be much larger than that for
intermediate generation (k1).
1.2
normalised concentration
Concentration versus log time curves
for reactant A, intermediate X, and
product P when the rate equations
are solved using the QSSA.
Values used for the rate constants
are the same as those used above.
QSSA reproduces the concentration
profiles well and is valid.
1.0
P
A
0.8
uss (τ)
0.6
v ss (τ)
wss (τ)
0.4
0.2
X
0.0
0.01
0.1
1
log τ
10
100
normalised concentration
1.2
u(τ)
v(τ)
w(τ)
1.0
0.8
k-1
P
A
0.6
X
0.4
0.2
0.0
0.01
k1
A
0.1
1
10
100
X
k2
P
Concentration versus log time curves
for reactant A, intermediate X and
product P when full set of coupled
rate equations are solved without
any approximation.
k-1 << k1, k2,,k1 and k-1 = k2 = 0.1
The concentration of intermediate is
high and it is present throughout much
of the duration of the experiment.
log τ
1.2
normalised concentration
Concentration versus log time curves
for reactant A, intermediate X and
product P when the Coupled rate
equations are solved using
the quasi steady state approximation.
The same values for the rate constants
were adopted as above.
The QSSA is not good in predicting
how the intermediate concentration
varies with time, and so it does not
apply under the condition where the
concentration of intermediate will be
high and the intermediate is long lived.
A
1.0
0.8
0.6
P
uss (τ)
v ss (τ)
wss (τ)
X
0.4
0.2
0.0
0.01
0.1
1
log τ
10
100
Macromolecule Formation : Polymerization.
H H
H H
C
C
C
H X
C
H X
n
Polymerization of vinyl halides
occurs via a chain growth addition
polymerization mechanism.
polymer
monomer
poly(vinyl chloride)
vinyl halide
X = Cl, vinyl chloride
3 step process :
• initiation
• propagation
• termination
Polymer = large molar mass molecule
(Macromolecule).
Chain initiation : highly reactive transient molecules or active centers (such
as free radicals formed.
Chain propagation : addition of monomer molecules to active chain end
accompanied by regeneration of terminal active site.
Chain termination : reaction in which active chain centers are destroyed.
We focus on radical addition polymerization . Kinetics of
polymerization quantified via QSSA .
Free Radical Addition Polymerization.
Initiation.
initiator
species
reactive free
radicals generated
ki = initiation rate
constant
kp = propagation
rate constant
kT = kTC + kTD
= termination rate
constant
ki
I →
2R ⋅
R ⋅ +M → M1 ⋅
active monomer
not kinetically
significant
macroradical
monomer
Propagation .
k
p
M j ⋅ + M →
M j +1 ⋅
assume that kp is independent
of chain length
k
p
M j +1 ⋅ + M →
M j+2 ⋅
Termination
TC
M n ⋅ + M m ⋅ k→
Pn + m
TD
M n ⋅ + M m ⋅ k→
Pn + Pm
termination via combination
termination via
disproportionation
Free Radical Addition Polymerization.
Analysis of the steady state kinetics.
Propagating macroradical
concentration will be small
since they are very reactive
so QSSA can be applied.
polymerization rate = rate of chain propagation
dm
RP = −
= rP = k P mx
dt
dx
≅0
dt
macroradical
concentration
dx
= rx formation − rx removal ≅ 0
dt
rx formation = rx removal
bimolecular propagation monomer concentration
rate constant
RP = rP = k P mx
1/ 2
 ri 
= kP 
 m
 2 kT 
ri can be left unspecified
depends on initiation
mechanism
ri = 2kT x 2
ri
x=
2 kT
initiation rate ri
ri = rT
rT = 2kT x 2
termination rate rT
2 macroradicals
disappear for each
incidence of termination
Application of QSSA. Kinetics of enzyme reactions.
Enzymes are very specific biological catalysts.
A catalyst is a substance that increases the rate of a
reaction without itself being consumed by the process.
• A catalyst lowers the Gibbs energy of activation ∆G ‡ by providing a different
mechanistic pathway by which the reaction may proceed. This alternative mechanistic
path enhances the rate of both the forward and reverse directions of the reaction.
• The catalyst forms an intermediate with the reactants in the initial step of the reaction
( a binding reaction), and is released during the product forming step.
• Regardless of the mechanism and reaction energetics a catalyst does not effect ∆H or
∆G of the reactants and products. Hence catalysts increase the rate of approach to
equilibrium, but cannot alter the value of the thermodynamic equilibrium constant.
energy
EA lowered
thermodynamics
unchanged
catalyst absent
∆EA
catalyst present
reactants
∆H
products
reaction coordinate
A reactant molecule acted upon
by an enzyme is termed a substrate.
The region of the enzyme where the
substrate reacts is called the active
site. Enzyme specificity depends on
the geometry of the active site and the
spatial constraints imposed on this region
by the overall structure of the enzyme
molecule.
Space filling models of
the two conformations of
the enzyme hexokinase.
(a) the active site is not
occupied. There is a cleft in
the protein structure that allows
the substrate molecule glucose
to access the active site.
(b) the active site is occupied.
The protein has closed around
the substrate.
CH2OH
H
H
O
H H
OH
OH
H OH
glucose
+ ATP
hexokinase
hexokinase
glucose
CH2OPO32H
H
O
OH H
H
+ ADP + H+
OH
OH
H OH
glucose 6- phosphate
Enzyme lock/key mechanism :
natural molecular recognition.
hexokinase
glucose
Chymotrypsin :
A digestive
enzyme .
substrate
active site
Natural
molecular
recognition in
action.
binding pocket
Mechanism of enzyme action.
Classification of enzymes.
Michaelis-Menten kinetics.
Michaelis-Menten rate
equation.
RΣ =
S+E
kc
ES
substrate
binding
E+P
ES complex
decomposition
to form products
enzyme/substrate
complex
1.2
RΣ
u
=
eΣ k c 1 + u
Ψ ≅1
u >> 1
1.0
Ψ=RΣ/kceΣ
Ψ=
kc eΣ c
KM + c
KM
Ψ
0.8
u << 1
c << K M
0.6
zero order
kinetics
RΣ ≅ kc eΣ
Ψ≅u
1st order
kinetics
0.4
RΣ ≅
kc
eΣ c
KM
Ψ
u
0.2
u
c >> K M
0.0
0
KM = Michaelis constant (mol dm-3)
kc = catalytic rate constant (s-1)
c = substrate concentration (mol dm-3)
eΣ = total enzyme concentration (mol dm-3).
10
20
30
40
50
u=c/KM This is an example of a complex
rate equation, where the reaction
rate varies with reactant concentration
in a non linear way.
Experimental rate equation : R = − ds = as
Σ
dt b + s
a, b = constants
Mechanism :
E+S
enzyme
ES
Proposed mechanism should
produce this rate law.
E+P
EP
enzyme/substrate
substrate complex
substrate
concentration
enzyme/product
complex
product
Rate equations
We analyse a simpler scheme which includes all the essentials.
p = [P]
s = [S]
E+S
k1
k2
k-1
e = [E]free
E+P
ES
k-2
x = [ES]
ds
= k1e s − k −1 x
dt
dx
−
= (k 2 + k −1 )x − k1e s − k − 2 e p
dt
dp
= k2 x − k−2e p
dt
−
What forms does the enzyme take?
• free enzyme E
Usually eΣ << s . Subsequent to mixing one has an
• bound enzyme ES
initial period during which x = [ES] builds up. We
then assume that the equilibrium concentration of ES
is rapidly attained and reaches a constant low value
eΣ = x + e
during the course of the reaction.
This requirement satisfies the QSSA.
total initial
enzyme concentration
QSSA
dx
= {k1s + k −1 + k 2 + k − 2 p}xSS − k1seΣ − k − 2 peΣ ≅ 0
dt
xSS =
(k1s + k −2 p )eΣ
k1s + k − 2 p + k −1 + k 2
RΣ = −
ds {k1k 2 s − k −1k − 2 p}eΣ
=
dt k1s + k − 2 p + k −1 + k 2
Let’s assume that measurement of the reaction rate
occurs during a time period when only a small percentage
(1-3%) of substrate is transformed to product.
p≅0
s ≅ s0
initial substrate
concentration
RΣ ≅ RΣ , 0
k1k 2 eΣ s0
k 2 eΣ s0
≅
=
k −1 + k 2 + k1s0 k −1 + k 2 + s
0
k1
initial rate
Michaelis
constant
mol dm-3
RΣ , 0
k +k
K M = −1 2
k1
kC = k 2
s-1
catalytic rate
constant
Fundamental kinetic
parameters : KM and
kC .
kC eΣ s0
=
K M + s0
Michaelis-Menten (MM)
equation for steady
state enzyme kinetics.
KM : enzyme/substrate
binding
kC : decomposition of
enzyme/substrate complex.
1
RΣ ,0
How can we evaluate KM and kC?
• Non linear least squares fitting to MM equation.
• Suitable linearization of MM equation.
1
K 1
1
= M
+
RΣ , 0 kC eΣ s0 kC eΣ
This has the same form
as the empirical rate
equation observed
experimentally.
Lineweaver-Burk
Lineweaver-Burk
Plot.
Plot.
S LB =
KM
kC eΣ
I LB =
1
kC eΣ
SLB
ILB
1
s0
unsaturated
enzyme kinetics
RΣ , 0
kC eΣ s0
dp
=
= k Σ eΣ =
dt
K M + s0
k C s0
kΣ =
K M + s0
1
KM
1
=
+
k Σ k C s0 k C
saturated
enzyme
kinetics
composite rate
constant
We consider two limiting behaviours.
• s0 << KM unsaturated enzyme kinetics ; not all active sites bound with substrate
• s0 >> KM saturated enzyme kinetics ; all active sites bound with substrate.
Case
CaseUS
US
s0
<< 1
KM
1
KM
>>
k C s0
kC
1
K
≅ M
k Σ k C s0
k
k Σ ≅ C s0
KM
kU =
(k k )k K k
kC
kk
= 1 2 = 1 −1 2 = 1 2
k
K M k −1 + k 2
1+θ
1+ 2
k −1
equilibrium constant
for ES formation
K1 =
θ=
k1
k −1
k2
k −1
We consider two
sub cases depending
on the value of θ .
1+θ ≅ θ =
θ >> 1 scenario.
k2
k −1
k 
kU ≅ K1k 2  −1  = k −1 K1 = k1
 k2 
RDS
k-1
k1
k2
When k2 >>k-1 , ES complex
decomposition to form
products is faster than ES
decomposition back to
reactants. The rds will
involve the rate of ombination
of E and S to form the ES
complex. The ES complex
will be short lived since it
does not accumulate to form
products.
RDS
ES
E+S
k-1
E+P
k2
k1
θ << 1 scenario.
1+θ ≅ 1
kU ≅ K1k 2
Have a fast pre-equilibrium
followed by a slow rate determining
decomposition of the ES complex
to form products.
ES
E+S
E+P
Case S
s0 >> K M
s0
>> 1
KM
KM
1
<<
k C s0
kC
1
1
≅
k Σ kC
k Σ ≅ kC = k 2
Case US
Saturated enzyme kinetics : all active sites
in enzyme bound with substrate.
Have slow rate determining breakdown of
ES complex to form products.
The detailed form of KM and kC in terms of fundamental rate
constants will depend on the nature of the mechanism.
E+S
k1
k-1
k-2
LB inverse plot will always pertain.
Pre equilibrium
KM
1
1
1
= +
+
kC
k1 K1k 2 K1 K 2 k3
ES complex
formation rds
ES
k2
Slow rate
determining
ES->EP
transformation
EP
k-3
E+P
eΣ
KM 1 1
=
+
RΣ , 0
k C s0 k C
Case S
Slow rate
determining
EP complex
decomposition.
k3
Pre equilibrium
1
1 1
1
= + +
kC k 2 k3 K 2 k3
EG
energy
gas phase
pathway
EA
gas phase
reactants
adsorbed
reactants
surface
catalysed
pathway
ED
ES
adsorbed
products
gas phase
products
Ej = activation
energy for step j.
Reaction intermediates stabilized via bonding
to catalytic surface sites.
G : gas phase reaction
A : adsorption
S : surface reaction
D : desorption
The Metal-Catalyzed Hydrogenation
of Ethylene H C CH (g) + H (g)
2
C2 H 4 ads
H
2
•
ads
2
reaction between
adsorbed species
H 3C
CH3 (g)
surface migration
•
H ads
dissociative
adsorption
H 2C • − CH 3 ads
•
H ads
Adsorption. Term used to describe the process whereby a molecule (the
adsorbate) forms a bond to a solid surface (an adsorbent).
number of sites occupied by adsorbate
NS
Fractional surface coverage θ
θ=
NΣ
total number of adsorption sites
When θ = 1, NS = NΣ and an
adsorbed monolayer is formed.
A (g)
The fractional coverage θ depends
on pressure of adsorbing gas phase species.
dynamic
This θ = θ (p) relationship is called an adsorption
equilibrium
isotherm.
A
surface
Langmuir
LangmuirAdsorption
AdsorptionIsotherm.
Isotherm.
ads
Simple approach to quantitatively describe an adsorption
process at the gas/solid interface.
Assumptions :
• solid surface is homogeneous and contains a number of equivalent sites, each
of which is occupied by a single adsorbate molecule
• a dynamic equilibrium exists between gas phase reactant and adsorbed species
• no interactions between adsorbed species
• adsorbed species localised, ∆ Hads is independent of coverage θ
associative adsorption
A (g) + S
kA
kD
A2 (g) + S
kA
kD
adsorption rate constant
Aads
desorption rate constant
2 Aads
K=
surface adsorption site
K measures affinity
of a particular
molecule for an
adsorption site.
kA
kD
dissociative adsorption
At equilibrium : RA = RD
k A p (1 − θ ) = k Dθ
Rate of adsorption :
R A = k A p (1 − θ )
pressure fractional coverage
Kp
θ=
1 + Kp
Langmuir adsorption
isotherm :
associative adsorption.
1 slope = 1/K
of vacant sites
Rate of desorption :
RD = k Dθ
fractional surface coverage
1
1
= 1+
θ
Kp
1
θ
1
p
A similar analysis can be done for dissociative adsorption.
Adsorption
rate
1
RD = k Dθ 2
θ
2
Desorption
rate
At equilibrium :
RA = k A p(1 − θ )
S=
RA = RD
k A p(1 − θ ) = k Dθ 2
1
K
2
Adsorption isotherm
for dissociative
adsorption.
θ2
kA
p = Kp
=
2
(1 − θ ) k D
θ=
Kp
1 + Kp
1
θ
= 1+
1
Kp
1
p
high p
limit :
monolayer
formed
Langmuir Adsorption Isotherm.
Kp >> 1
1 + Kp ≅ Kp
θ →1
Surface coverage θ
1.0
Low p limit :
Henry Law
Kp << 1
1 + Kp ≅ 1
θ ≅ Kp
0.8
K large
0.6
K = 10
K=1
K = 0.1
0.4
K small
0.2
0.0
0.0
0.5
1.0
1.5
P/atm
Kp
θ=
1 + Kp
2.0
Kinetics of surface reactions.
Assume that gaseous reactant decomposes when it is adsorbed.
Reaction
rate
R = kθ
Surface coverage of adsorbed gas
kKp
R=
1 + Kp
Kp
θ=
1 + Kp
We can consider two limits.
High pressures.
Kp >> 1
R≅k
Rate independent of
Gas pressure p
Zero order kinetics.
Adsorption rate very
large when p is high.
Decomposition step rds.
Surface coverage
related to gas
pressure p via
Langmuir adsorption
isotherm
Low pressures.
Kp << 1
R ≅ kKp
Rate depends
linearly on gas
Pressure p
First order kinetics.
Adsorption process
is rate determining
when p is low.
Decomposition is fast.
Adsorption of a gas on a solid is
an exothermic process : ∆ Hads is
negative.
EA
A(g)
Both adsorption and desorption
processes follow the Arrhenius
equation.
adsorption pre-exponential factor
 E 
k A = AA exp − A 
 RT 
 E 
k D = AD exp − D 
 RT 
ED
∆Hads
activation energy for
adsorption
activation energy for
desorption
temperature (K)
desorption pre-exponential factor
R = gas constant = 8.314 J mol-1 K-1
Aads
k A AA
 ∆H ads 
=
exp −
K=

k D AD
 RT 
∆H ads = E A − ED
entropy of adsorption
How is ∆Hads measured ?
0
0
0
∆Gads
= − RT ln K = ∆H ads
− T∆S ads
entalphy of adsorption
Gibbs energy of adsorption
0
0
∆H ads
∆S ads
ln K = −
+
RT
R
0
∆H ads
 ∂ ln K 
 =

RT 2
 ∂T θ
constant surface coverage
Langmuir adsorption assumed
∆H ads
 ∂ ln p 
=
−


RT 2
 ∂T θ
0
1−θ
 θ 
ln K + ln p = ln 

1 − θ 
 ∂ ln K   ∂ ln p 
 =0
 +

T
T
∂
∂
θ
θ 

 ∂ ln K 
 ∂ ln p 

 = −

∂
∂
T
T
θ
θ


constant surface
coverage
0
  p1 
∆H ads
ln  =
R
  p2  θ
1 1 
 − 
T1 T2 
1.0
T2
θ
Kp =
θ
p1, p2, T1 and T2 can
be measured so
∆Hads can be determined.
T1
0.5
0.0
0
p2
1
p1
p
2
Chemical Kinetics.
Lectures 5-6.
Microscopic theory of chemical
reaction kinetics.
Temperature effects in chemical kinetics.
•
Chemical reactions are activated processes : they require an
energy input in order to occur.
• Many chemical reactions are activated via thermal means.
• The relationship between rate constant k and temperature T is
given by the empirical Arrhenius equation.
• The activation energy EA is determined from experiment, by
measuring the rate constant k at a number of different
temperatures. The Arrhenius equation
 EA 
k = A exp −
is used to construct an Arrhenius plot
 RT 
of ln k versus 1/T. The activation energy
Pre-exponential
is determined from the slope of this plot.
factor
 d ln k 
 d ln k 
 = RT 2 
E A = − R

 dT 
 d (1 / T ) 
ln k
Slope = −
1
T
EA
R
Microscopic theories of
chemical reaction kinetics.
•
•
•
•
A basic aim is to calculate the rate constant for a chemical reaction from
first principles using fundamental physics.
Any microscopic level theory of chemical reaction kinetics must result in
the derivation of an expression for the rate constant that is consistent
with the empirical Arrhenius equation.
A microscopic model should furthermore provide a reasonable
interpretation of the pre-exponential factor A and the activation energy EA
in the Arrhenius equation.
We will examine two microscopic models for chemical reactions :
– The collision theory.
– The activated complex theory.
•
The main emphasis will be on gas phase bimolecular reactions since
reactions in the gas phase are the most simple reaction types.
References for Microscopic Theory
of Reaction Rates.
• Collision Theory.
– Atkins, de Paula, Physical Chemistry 7th
edition, Chapter 27, Section. 27.1, pp.944951.
• Activated Complex Theory.
– Atkins, de Paula, Physical Chemistry 7th
edition, Chapter 27, Section.27.4-27.5, pp.
956-961.
Collision theory of bimolecular
gas phase reactions.
• We focus attention on gas phase reactions and assume
that chemical reactivity is due to collisions between molecules.
• The theoretical approach is based on the kinetic theory of gases.
• Molecules are assumed to be hard structureless spheres. Hence the
model neglects the discrete chemical structure of an individual
molecule. This assumption is unrealistic.
• We also assume that no interaction between molecules until contact.
• Molecular spheres maintain size and shape on collision. Hence the
centres cannot come closer than a distance δ given by the sum of
the molecular radii.
• The reaction rate will depend on two factors :
• the number of collisions per unit time (the collision frequency)
• the fraction of collisions having an energy greater than a certain
threshold energy E*.
Simple collision theory : quantitative aspects.
k
A( g ) + B ( g ) 
→
Products
Hard sphere reactants
Molecular structure and
details of internal motion
such as vibrations and rotations
ignored.
Two basic requirements dictate a collision event.
• One must have an A,B encounter over a sufficiently short distance to allow
reaction to occur.
• Colliding molecules must have sufficient energy of the correct type to
overcome the energy barrier for reaction. A threshold energy E* is required.
Two basic quantities are evaluated using the Kinetic Theory of gases :
the collision frequency and the fraction of collisions that activate molecules
for reaction.
To evaluate the collision frequency we need a mathematical way to define
whether or not a collision occurs.
A
The collision cross section σ
defines when a collision occurs.
σ = πδ = π (rA + rB )
2
2
rA
δ
rB
Effective collision
diameter
B
δ = rA + rB
Area = σ
The collision cross section for two molecules can be
regarded to be the area within which the center of the
projectile molecule A must enter around the target molecule B
in order for a collision to occur.
Cross sectional
Area of disc
A
σ =π δ 2
B
b≤δ
δ = rA + rB
b≤δ
b >δ
Criterion for
Successful collision
Collision
possible
b≤δ
Collision
impossible b > δ
b >δ
Maxwell-Boltzmann velocity
Distribution function
 m 
F (v) = 4π v 

2
π
k
T
B 

3/ 2
2
F (v)
Gas molecules exhibit
a spread or distribution
of speeds.
 mv 2 
exp −

2
k
T
B 

• The velocity distribution curve
has a very characteristic shape.
• A small fraction of molecules
move with very low speeds, a
small fraction move with very
high speeds, and the vast majority
of molecules move at intermediate
speeds.
• The bell shaped curve is called a
Gaussian curve and the molecular
speeds in an ideal gas sample are
Gaussian distributed.
• The shape of the Gaussian distribution
curve changes as the temperature
is raised.
• The maximum of the curve shifts to
higher speeds with increasing
temperature, and the curve becomes
broader as the temperature
increases.
•
A greater proportion of the
gas molecules have high speeds
at high temperature than at
low temperature.
v
The collision frequency is computed via the kinetic
Theory of gases.
We define a collision number (units: m-3s-1) ZAB.
Mean relative velocity
Units: m2s-1
Z AB = π δ 2 n A nB vr
nj = number density of molecule j (units : m-3)
δ = rA + rB
Mean relative velocity evaluated via kinetic theory.
Average velocity of a gas molecule
MB distribution of velocities
enables us to statistically
estimate the spread of
molecular velocities in a gas
∞
v = ∫ v F (v ) dv
0
m 
2 
F (v) = 4π v 

2
π
k
T
B 

3/ 2
Maxwell-Boltzmann velocity
Distribution function
 mv 2 
exp −

2
k
T
B 

Some maths !
8k B T
v =
πm
Mass of
molecule
We now relate the average velocity to the mean
relative velocity.
If A and B are different molecules then
Hence the collision number
between unlike molecules
can be evaluated.
vr =
Collision frequency
factor
2
+ vB
2
vj =
πµ
Reduced mass
Z AB = π δ n A nB vr
 8k T 
Z AB = n A nBσ  B 
π µ 
= Zn A nB
vA
8k B T
2
1/ 2
vr =
µ=
m A mB
m A + mB
For collisions between like molecules
The number of collisions per unit
time between a single A
molecule and other A
Z A = n Aσ
molecules
Total number of collisions
between like molecules
We divide by 2 to ensure
That each A,A encounter
Is not counted twice.
8k B T
π mj
vr = 2 v
1/ 2
 8k T 
2 B 
π m A 
1/ 2
Z AA
2 2  8k B T 
Z n
= A A =
nA σ 

2
2
π
m
A

E*
Molecular collision is
effective only if
translational energy
of reactants is
greater than some
threshold value.
Fraction of molecules
with kinetic energy
greater Than some
minimum Threshold
value ε*
 ε* 
F (ε > ε *) = exp −

 k BT 
• The simple collision theory expression for the reaction rate R between unlike
molecules
R=−
 ε* 
dn A
= Zn A nB exp −

dt
k
T
 B 
• The more usual rate expression for a
bimolecular reaction between A and B is
• We introduce molar variables
E* = N Aε *
a=
Avogadro
constant
nA
NA
b=
dn A
da
= NA
dt
dt
nB
NA
R=−
1/ 2
 8k T 
Z =σ  B 
π µ 
da
= kab
dt
• Hence the SCT rate expression becomes
R=−
da
 E *
= ZN A ab exp −
dt
 RT 
• The bimolecular rate constant for
collisions between unlike molecules is given by
• Similarly for bimolecular collisions
between like molecules
1/ 2
1/ 2


 k BT 
k
T
8
*
E
 E *
 Collision k = N σ

B
exp
−
=
2
exp
k
N
σ



A
A 
− RT 
 RT  Frequency
Both of these
π
µ
π
m




expressions are
factor
 E *
 E *
similar to the
= z AB exp −
= z AA exp −
 RT 
Arrhenius equation.
 RT 
We compare the results of SCT with the empirical Arrhenius eqn.
In order to obtain an interpretation of the activation energy and
Pre-exponential factor.
1/ 2
k T 
 E *
k = 2 N Aσ  B  exp −
 RT 
π m 
 E *
= z AA exp −
 RT 
A,A encounters
 EA 
k = Aobs exp −

 RT 
Aobs = z AA = A' ' T A
8k B
• SCT predicts
A' ' = 2 N Aσ
πm
that the pre-exponential
factor should depend on temperature.
• The threshold energy and the
activation energy can also be
compared.
• Activation energy exhibits
a weak T dependence.
A,B encounters
Aobs = z AB = A' T A
A' = N Aσ
d ln k
E
= A2
dT
RT
Arrhenius
1/ 2
 8k T 
 E *
k = N Aσ  B  exp −
 RT 
πµ 
 E *
= z AB exp −
 RT 
8k B
πµ
Pre-exponential
factor
SCT
d ln k E * + RT 2
=
dT
RT 2
EA = E * +
RT
2
EA ≅ E *
SCT : a summary.
•
•
•
•
•
•
•
•
The major problem with SCT is that the threshold energy E* is very
difficult to evaluate from first principles.
The predictions of the collision theory can be critically evaluated by
comparing the experimental pre-exponential factor with that computed
using SCT.
We define the steric factor P as the ratio between
exp
calc
the experimental and calculated A factors.
We can incorporate P into the SCT
expression for the rate constant.
 E *
exp
=
k
Pz
AB
For many gas phase reactions
− RT 
P is considerably less than unity.
 E *
Typically SCT will predict that Acalc will be in
k = Pz AA exp −
 RT 
the region 1010-1011 Lmol-1s-1 regardless of
the chemical nature of the reactants and products.
What has gone wrong? The SCT assumption of hard sphere collision
neglects the important fact that molecules possess an internal structure.
It also neglects the fact that the relative orientation of the colliding
molecules will be important in determining whether a collision will lead to
reaction.
We need a better theory that takes molecular structure into account. The
activated complex theory does just that .
P=A
A
Summary of SCT.
A,B encounters
1/ 2
 8k T 
 E *
k = N Aσ  B  exp −
 RT 
πµ 
 E *
= z AB exp −
 RT 
A,A encounters
1/ 2
k T 
 E *
k = 2 N Aσ  B  exp −
 RT 
π m 
 E *
= z AA exp −

 RT 
Transport
property
Energy criterion
 E *
k = Pz AB exp −

 RT 
 E *
k = Pz AA exp −

 RT 
Steric factor
(Orientation requirement)
Weaknesses:
• No way to compute P from molecular
parameters
• No way to compute E* from first principles.
Theory not quantitative or predictive.
Strengths:
•Qualitatively consistent with observation
(Arrhenius equation).
• Provides plausible connection between
microscopic molecular properties and
macroscopic reaction rates.
• Provides useful guide to upper limits for rate
constant k.
A + BC → [ ABC ] → AB + C
*
r
EA
s
EA
∆U 0
Activated complex
Transition state
energy
Progress of a chemical reaction can be
expressed in terms of a plot of energy
versus reaction co-ordinate.
The reaction coordinate may be described
in terms of changes in particular bond lengths
since these will vary as the reaction progresses.
∆G*
products
reactants
Reaction coordinate
Activated Complex
or Transition State
The transition state or activated complex is a high energy
species of fleeting (ca fs lifetime, 1fs = 10-15s) existence. Its structure
has features both of the reactants and the products.
1999 Nobel Prize in Chemistry awarded to Ahmed Zewail
from Caltech for his studies of transition states of chemical
reactions by femtosecond spectroscopy using laser technology.
Experimental study of very short timescales is called
femtochemistry.
Transition states can have different geometries.
A (g) + B(g)
X*
ν*
Products
Basic assumption : activated complex
X* is treated as a thermodynamic
quantity in thermodynamic
equilibrium with the reactants.
Kc*
Bimolecular reaction
frequency of decomposition
of activated complex
Equilibrium constant for
reactant/transition state
transformation.
x∗
∗
Kc =
ab
R =ν x
∗ ∗
Reaction rate R depends on the frequency of
decomposition and concentration of activated
complexes.
bimolecular reaction
∗
R = ν K c ab
∗
R = kab
∗
x ∗ = K c ab
Reaction molecularity m : number of
molecules which come together to form
the activated complex.
m = 1 : unimolecular reaction
m = 2 : bimolecular reaction .
rate
constant
k = ν ∗Kc
∗
This fundamental expression must now be
evaluated.
For an ideal gas or for reactions in solution :
∗
K C = K ∗ (c 0 ) ∆n*
thermodynamic
equilibrium constant
k = ν KC
∗
h = Planck’s constant = 6.63x10-34Js
kB = Boltzmann constant = 1.38x10-23 JK-1
For T = 298 K
ν ≅ 6 × 10 s
∗
12
−1
∆ n ∗ = 1 − 2 = −1
Change in # moles
for reactant /TS
transformation
We assume that the TS
decomposes with a
frequency given by:
k BT
∗
ν =
h
1 molL-1
∗
K
= ν  0
c
∗
∗



KC
∗
K∗
= 0
c
k has units Lmol-1s-1
Not all transition states go and form
products.
 K∗ 
k = κ ν K C = κ ν  0 
c 
k T 
= κ  B 0 K ∗
 hc 
∗
∗
This is of the correct
order of magnitude for
a molecular vibration
frequency.
κ = transmission coefficient
0<κ<1
∗
We relate K* to the Gibbs energy of activation ∆G*
 ∆G
 k BT 
k = κ  0  exp −
 hc 
 RT
0∗
 ∆S 0∗ 
 ∆H 0∗ 
 k BT 
k = κ  0  exp 
 exp −

R
RT
 hc 




 EA 
k = A exp −
 RT 
= − RT ln K ∗
0∗ 

∆G
∗
K = exp −

 RT 



Eyring equation : fundamental ACT result.
∆G
0∗
∆G
0∗
= ∆H
0∗
− T ∆S
enthalpy of
activation
We obtain a useful
interpretation for
activation energy and
pre-exponential factor.
0∗
entropy of
activation
0∗ 

 ∆H 0∗ 
∆S
 k BT 
k = κ  0  exp 
 exp −

R
RT
 hc 




 d ln k 
E A = RT 2 

 dT 
internal energy of activation

∆S 0∗ 
 EA 
 k BT 
exp
k = κ  0  exp 2 +
−


R 
 RT 
 hc 

E A = ∆U 0∗ + RT
∆H 0∗ = ∆U 0∗ + P∆V 0∗
condensed phases
volume of activation
m = 2 bimolecular reaction
∆V 0∗ ≅ 0
P∆V 0∗ = ∆n∗ RT = (1 − m )RT = − RT
ideal gases
pre-exponential factor A
reaction
molecularity
bimolecular gas
phase reaction
∆H 0∗ = E A − mRT
m = 1, condensed phases, unimolecular
gas phase reactions
m = 2, bimolecular gas phase reactions
0∗
 k BT 


S
∆
 EA 


km = κ
exp m +
 exp −

 h c 0 m −1 
R
RT





m = molecularity
( )
∆S0* explained in
terms of changes in translational,
rotational and vibrational
degrees of freedom on going from
reactants to TS.
0∗
 k BT 


S
∆


A =κ
exp m +

 h c 0 m −1 
R




( )
pre-exponential factor related to entropy
of activation (difference in entropy between
reactants and activated complex
( )
steric factor
S =−
EA
R
1
T
ln A
0∗
 k BT 


S
∆
0∗
 exp m +
A = PZ = κ 
P
≅
1
∆
S
≅0

m
1
−
 h c0

R 



P < 1 ∆S 0∗ negative
collision theory
ln k
P > 1 ∆S 0∗ positive
TS more ordered
than reactants
TS less ordered than
reactants