Kinetika Kimia
(Chemical Kinetics)
Jaslin Ikhsan, Ph.D.
Course Summary.
• Dependence of rate on concentration.
• Experimental methods in reaction
kinetics.
• Kinetics of multistep reactions.
Recommended reading.
1. P.W. Atkins, Kimia Fisika (terjemahan), jilid 2, edisi keempat,
2. P.W. Atkins, Physical Chemistry, 5th ed., Oxford:
3. Gordon M. Barrow, Physical Chemistry
4. Arthur M. Lesk, Introduction to Physical Chemistry
Oxford University Press, 1994
Jakarta: Erlangga, 1999.
Chemical Kinetics.
Lecture 1.
Chemical reaction kinetics.
reactants products
• 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?
)
(
2
)
(
)
(
22
g
I
g
HI
g
H
+
→
• 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 willBasic ideas in reaction kinetics.
• Chemical reaction kinetics deals with the rate of velocity of chemical reactions.
• We wish to quantify
– The velocity at which reactants are transformed to products
– The detailed molecular pathway by which a reaction proceeds (the reaction mechanism).
• 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.
Reaction Rate.
• What do we mean by the term reaction rate?
– The term rate implies that something changes with respect to something else.
• How may reaction rates be determined ?
– 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.
[ ] [ ]
dt
P
d
dt
R
d
R
Σ
=
−
=
Product
concentration Reactant
concentration
Net reaction rate Units : mol dm-3 s-1
[R]t [P]t
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 Jwith respect to changes in time. The magnitude of the reaction rate changes as the reaction proceeds.
[ ]
[ ]
dt
J
d
t
J
R
J t
J
J
υ
υ
1
lim
1
0
∆
=
∆
=
→
∆
[ ]
[ ]
[
]
dt O H d dt
O d dt
H d R
g O H g
O g H
2 2
2
2 2
2
2 1 2
1
) ( 2
) ( )
( 2
= −
= −
=
→ +
•
Note : Units of rate :- concentration/time , hence RJ has units mol dm-3s-1 .υ
J denotes the stoichiometric coefficient of species J. If J is a reactantυ
J 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.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).
Products
→
+
kyB
xA
[ ]
[ ] [ ] [ ]
α βB
A
k
dt
B
d
y
dt
A
d
x
R
=
−
1
=
−
1
=
rate constant k
stoichiometric coefficients
empirical rate
equation (obtained from experiment)
α, β = reaction orders for the reactants (got experimentally)
Rate equation can not in general be inferred from the stoichiometric equation for the reaction.
Log [A]
Slope =
α
Log R
Log R
Log [B]
Cl
Br
I
X
HX
X
H
,
,
2
2 2=
→
+
Different rate equations
imply different mechanisms.
[ ] [ ][ ]
[
] [ ][ ]
[
]
[ ]
[
] [ ][ ]
1/22 2 2 2 2 2 / 1 2 2 2 2 2 2 2 2
2
1
2
2
Cl
H
k
dt
HCl
d
R
HCl
Cl
H
Br
HBr
k
Br
H
k
dt
HBr
d
R
HBr
Br
H
I
H
k
dt
HI
d
R
HI
I
H
=
=
→
+
′
+
=
=
→
+
=
=
→
+
•
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.
Zero order kinetics.
The reaction proceeds at the same rate R
regardless of concentration.
[ ]
0A
R
∝
R
Rate equation :
units of rate constant k : mol dm-3 s-1
0
when
0
=
=
=
−
=
t
a
a
k
dt
da
R
]
[
A
integrate using initial condition
0 2
/ 1 0
2 / 1
0 2
/ 1
2
2
when
a
k
a
a
a
t
∝
=
=
=
τ
τ
τ
half life :
0
)
(
t
kt
a
a
=
−
+
a
t
slope = -k
0
a
diagnostic
plot
2 / 1
τ
0
a
k slope
2 1
products
→
kA
First order kinetics.
First order differential
rate equation.
[ ]
dtA d rate = − t
Initial
concentration a0
ka
dt
da
=
−
Initial condition
0
0
a
a
t
=
=
Solve differential equation
Separation of variables
[ ]
kt
a
e
a
t
a
(
)
=
0 −kt=
0exp
−
For a first order reaction the relationship:
[ ]
[ ]
tt
t t
A
k
rate
A
rate
=
∝
)
(
)
(
is valid generally for any time t.
k is the first order
rate constant, units: s
-12 / 1 2 2 / 1 0 2 / 1 = = = = u a a t θ θ τ
k
k
693
.
0
2
ln
2 /1
=
=
τ
Half life
τ
1/2First order kinetics.
1 −
→
s
k
[ ]
[ ]
kt
a
t
a
u
kt
a
e
a
t
a
kt=
−
=
=
−
=
=
−θ
θ
exp
)
(
exp
)
(
0 0 0Mean lifetime of reactant molecule
1
=
u
5
.
0
=
u
25
.
0
=
u
125
.
0
=
u
0
a a u =
2
/
0 2
/
1
a
a
t
=
τ
=
2 / 1
τ
k
k
693
.
0
2
ln
2 /
1
=
=
τ
First order kinetics: half life.
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.
half life independent of
initial reactant concentration
0
1st order kinetics
2nd order kinetics
θ
θ
−−
= =
e u
e a t
a kt
) (
)
( 0
)
(
θ
u
θ θ
+ =
+ =
1 1 )
(
1 ) (
0 0
u
t ka a t
a
)
(
θ
u
kt
=
θ
ka
t
0
=
dm
3mol
-1s
-1Second order kinetics: equal
reactant concentrations.
2
A
→
kP
0 2
0
a
a
t
ka
dt
da
=
=
=
−
separate variables
integrate
0
1
1
a
kt
a
=
+
a
1
slope = k
t
2
0 2
/ 1
a a t =
τ
=half life
( )
t
ka
a
t
a
0 0
1
+
=
2 / 1
τ
↑ ↓
∝ =
0 2
/ 1
0 2
/ 1
0 2
/ 1
1 1
a as a ka
τ τ τ
rate varies as
square of reactant
concentration
0
1st and 2nd order kinetics : Summary .
Reaction Differential rate equation Concentration variation with time Diagnostic Equation Half Life Products 1 → k A a k dt da 1 =− a t a
[
kt]
1 0 exp ) ( − = 0 1 ln ) (
lna t =−kt+ a
1 2 / 1 2 ln k = τ Products 2A→k2
2 2a k dt da = − t a k a t a 0 2 0 1 ) ( + = 0 2 1 ) ( 1 a t k t
a = + 1/2 2 0
1
a k
=
τ
Slope = - k1
Slope = k2
n th order kinetics: equal reactant
concentrations.
nA
P
k
→
11
− na
separate variables integrate 00
a
a
t
ka
dt
da
n=
=
=
−
(
)
10 1
1
1
1
−−
=
−
+
nn
a
kt
n
a
(
n)
k slope = −11
≠
n
n = 0, 2,3,…..
rate constant k
obtained from slope
t
Half life
(
)
10 1 2 / 1
1
1
2
− −−
−
=
n nka
n
τ
↑
↑
<
↑
↓
>
∝
− 0 2 / 1 0 2 / 1 1 0 2 / 11
1
a
as
n
a
as
n
a
nτ
τ
τ
(
)
(
)
01 2
/
1 1 ln
1 1 2
ln
ln n a
k n n − − − − = − τ 2 / 1
ln
τ
(
−1)
−
= n
slope
reaction order n determined
from slope
0
Summary of kinetic results.
P
nA
→
k2 0 0 2 / 1 0 a a t a a t = = = = τ Rate equation
Reaction
Order
dt
da
R
=
−
expression
Integrated
Units of k
Half life
τ
1/20
k
a( )
t = −kt+a0mol dm
-3
s
-1k
a
2
01
ka
( )
ktt a a = 0 ln
s
-1 k 2 ln2
2ka
( )
0 1 1 a kt t
a = +
dm
3mol
-1s
-10
1
ka
3
3ka
( ) 20 2 1 2 1 a kt t
a = +
dm
6mol
-2s
-12 0
2 3
ka
n
ka
n ( ) 10 1 1 1 1 −
− = − + n
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 ?
3. Derive the rate equation for reaction whose orders are:
(a). one-half, (b). three and a-half, (c). 4, and (d). n !
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
Chemical Kinetics
Lecture 2.
Kinetics of more complex
reactions.
Jaslin Ikhsan, Ph.D.
Kimia FMIPA
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
P
B
A
+
→
kSecond order kinetics:
Unequal reactant concentrations.
rate equation
kab
dt
dp
dt
db
dt
da
R
=
−
=
−
=
=
initial conditions
0 0
0 0
0
a
a
b
b
a
b
t
=
=
=
≠
dm
3mol
-1s
-1integrate using
partial fractions
( )
slope = k
b
a
F
,
( )
kt
a
a
b
b
a
b
b
a
F
=
−
=
0 0
0 0
ln
1
,
Consecutive Reactions .
A
→
k1X
→
k2P
•Mother / daughter radioactive
decay. Mass balance requirement:
1 4 2 1 3 1 214 214 218
10
6
10
5
×
− −=
×
− −=
→
→
s
k
s
k
Bi
Pb
Po
x
a
a
p
=
0−
−
The solutions to the coupled equations are :
3 coupled LDE’s define system :
[
]
[
]
[
]
{
}
[
]
{
[
k t]
[
k t]
}
k k a k t k a a t p t k t k k k a k t x t k a t a 2 1 1 2 0 1 1 0 0 2 1 1 2 0 1 1 0 exp exp exp ) ( exp exp ) ( exp ) ( − − − − − − − = − − − − = − =
x
k
dt
dp
x
k
a
k
dt
dx
a
k
dt
da
2 2 1 1=
−
=
−
=
Consecutive reaction : Case I.
Intermediate formation fast, intermediate decomposition slow.
1 2
1
2 1
k k
k k
<< << =
κ
Case I .
TS II
TS I
P
X
A
→
k1
→
k2∆
G
I‡∆
G
II‡
energy
I : fast
II : slow
rds
∆
G
I‡<<
∆
G
II‡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
0 0
0
a
p
w
a
x
v
a
a
u
=
=
=
Initial reactant A more
reactive than intermediate X .
τ = k1t
0 2 4 6 8 10
No
rm
al
is
e
d
c
o
nc
e
n
tr
at
io
n
0.0 0.2 0.4 0.6 0.8 1.0 1.2
u = a/a0 v = x/a0 w = p/a0
κ=k
2/ k1 = 0.1
Reactant A
Product P
Intermediate X
P
X
A
→
k1
→
k2 2 11
2 1
k k
k k
<< << =
κ
Concentration of intermediate significant over time course of reaction.
Consecutive reactions Case II:
Intermediate formation slow, intermediate decomposition fast.
1 2
k
k
=
κ
P
X
A
→
k1
→
k2key parameter
Case II .
Case II .
1 2
1
2 1
k k
k k
>> >> =
κ
Intermediate X fairly reactive.
[X] will be small at all times.
energy
TS I
A
X
P
TS II
reaction co-ordinate
∆
G
I‡∆
G
II‡P
X
A
→
k1
→
k2I : slow rds
II : fast
∆
G
I‡>>
∆
G
II‡P
X
A
→
k1
→
k2τ = k1t
0 2 4 6 8 10
no
rm
al
is
e
d
c
o
nc
e
n
tr
at
io
n
0.0 0.2 0.4 0.6 0.8 1.0 1.2
u=a/a0 v =x/a0 w=p/a0
κ = k2/k1 = 10
Reactant A
Product P
Intermediate X
τ = k1t
0.0 0.2 0.4 0.6 0.8 1.0
n
o
rm
al
is
e
d
c
o
nc
e
n
tr
ati
o
n
0.0 0.2 0.4 0.6 0.8 1.0 1.2
u=a/a0
v=x/a0
w=p/a0
A P
X
Intermediate concentration is approximately constant after initial induction period.
1 2
1
2 1
k k
k k
>> >> =
κ
Case II .
Case II .
Rate Determining Step
2
1
k
k
>>
P
X
A
→
k
1
→
k
2Fast
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 .
1
2
k
k
>>
P
X
A
→
k
1
→
k
2Slow
Fast
Rate Determining Step
•
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.
Parallel reaction mechanism.
Y
A
X
A
k k→
→
2 1• We consider the kinetic analysis of a concurrent or parallel reaction scheme which is often met in real situations.
• A single reactant species can form two distinct products.
We assume that each reaction exhibits 1st order
kinetics.
k1, k2 = 1st order rate constants
We can also obtain expressions for the product concentrations x(t) and y(t).
(
)
[
]
(
)
[
]
(
)
[
]
{
k k t}
k k a k t x dt t k k a k t x t k k a k a k dt dx t 2 1 2 1 0 1
0 1 2
0 1 2 1 0 1 1 exp 1 ) ( exp ) ( exp + − − + = + − = + − = =
∫
• Initial condition : t= 0, a = a0 ; x = 0, y = 0 .• Rate equation:
(
k k)
a k a a k a k dt daR = − = 1 + 2 = 1 + 2 = Σ
[
k
t
]
a
[
(
k
k
)
t
]
a
t
a
(
)
=
0exp
−
Σ=
0exp
−
1+
2(
)
[
]
(
)
[
]
(
)
[
]
{
k k t}
k k a k t y dt t k k a k t y t k k a k a k dt dy t 2 1 2 1 0 2
0 1 2
0 2 2 1 0 2 2 exp 1 ) ( exp ) ( exp + − − + = + − = + − = =
∫
2 1 2 / 1 2 ln 2 ln k k k = +=
Σ
τ
• Half life:
2 1 ) ( ) ( k k t y t x Lim
t→∞ =
Final product analysis
yields rate constant ratio. • All of this is just an extension of simple
Parallel Mechanism: k1 >> k2
τ = k1t
0 1 2 3 4 5 6
no
rm
al
is
ed
c
o
nc
en
tr
a
ti
o
n
0.0 0.2 0.4 0.6 0.8 1.0
u(τ) v(τ) w(τ)
κ = k2
//k1
a(t)
x(t)
y(t)
1 . 0 =
κ
( )∞ =0.9079
v
( )∞ =0.0908
w
( )
( ) 0.0908 9.9989 9079
.
0 =
≅ ∞ ∞
w v
( )
) ( 10
1 . 0
2 1
1 2
∞ ∞ = =
= =
w v k
k k k
κ Computation
Parallel Mechanism: k2 >> k1
τ = k1t
0 1 2 3 4 5 6
no
rm
al
is
e
d
c
o
nc
e
n
tr
at
io
n
0.0 0.2 0.4 0.6 0.8 1.0
u(τ) v(τ) w(τ)
a(t)
x(t) y(t)
10 =
κ
0909 . 0 ) (∞ ≅
v
9091 . 0 ) (∞ ≅
w
( )
) ( 1
. 0
10
2 1
1 2
∞ ∞ = =
= =
w v k
k k k
κ
( )
( ) 0.9091 0.0999 0909
.
0 =
≅ ∞ ∞
w v
Reaching Equilibrium on the
Macroscopic and Molecular Level
N
2O
4NO
2N
2O
4 (g)2 NO
2 (g)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.
[
]
[
]
↓ ↑ t t O N NO 4 2 2 Equilibrium state Kinetic regime NO2N2O4
concentration
N
2O
4 (g)2 NO
2 (g)colourless brown
[
]
[
N
O
]
eqeqNO
4 2 2 Concentrations vary with time Concentrations time invariant Kinetic analysis.[
]
[
]
2 2 4 2 NO k R O N k R ′ = = s r Equilibrium:[
]
[
]
[
]
[
]
K k k O N NO NO k O N k R R eq eq eq = ′ = ′ = = 4 2 2 2 2 2 4 2 s r ∞ → t ∞ → t time Equilibrium constantA B k
k'
First order reversible reactions :
understanding the approach to chemical equilibrium.
Rate equation
b k ka dt
da = − + ′ 0 1 0
1 = = = = + v u v u
τ
Rate equation in normalised form Initial condition
0
0
=
0=
=
a
a
b
t
θ
τ
+
=
1
+
1
u
d
du
Mass balance condition
Solution produces the concentration expressions
0
a
b
a
t
+
=
∀
( )
{
[ ]
}
( )
{
[ ]
τ}
θ θ τ τ θ θ τ − − + = − + + = exp 1 1 exp 1 1 1 v uIntroduce normalised variables.
(
)
k k t k k a b v a a u ′ = ′ + = = =τ
θ
00
( ) ( )
( )
+[ ]
[ ]
− − − = = τ θ τ θ τ τ τ exp 1 exp 1 u v QFirst order reversible reactions: approach to equilibrium.
τ = (k+k')t
0 2 4 6 8
co
n
c
e
n
tr
a
ti
o
n
0.0 0.2 0.4 0.6 0.8 1.0
u (τ) v (τ)
Equilibrium Kinetic
regime
Reactant A
Product B
(
)
( )
( )
∞
∞
=
∞
→
=
u
v
Q
τ = (k+k')t
0 2 4 6 8
Q(
τ
)
0 2 4 6 8 10 12
10
=
θ
Equilibrium Q = K = θ
Understanding the difference between reaction quotient Q and
Equilibrium constant K.
Approach to Equilibrium Q < θ
( )
( )
∞ ∞ =u v K
( ) ( )
( )
[ ]
[ ]
− +
− −
= =
τ θ
τ θ
τ τ τ
exp 1
exp 1
u v Q
k
k
K
Q
t
′
=
=
→
∞
C H H
C H
X
C H
H C H
X n
Chemical Kinetics.
Lecture 3/4.
Application of the Steady State
Approximation: Macromolecules
• Detailed mathematical analysis of complex 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
leads to a considerable simplification in the kinetic analysis.
Quasi-Steady State Approximation.
QSSA
P
X
A
→
k1
→
k2Consecutive reactions k1 = 0.1 k2 .
τ = k1t
0.0 0.5 1.0
N
o
rmalised concentration
0.0 0.5 1.0
u(τ)
v(τ) w(τ)
intermediate X induction
period
A
P
concentration approx. constant
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
removal X
formation X
removal X
formation X
R
R
R
R
dt
dx
=
≅
−
Consecutive reaction mechanisms.
A X P
k1
k-1
k2 Step I is reversible, step II is
Irreversible.
Coupled LDE’s can be solved via Laplace Transform or other methods.
(
)
v d dw v u d dv v u d du φ τ φ κ τ κ τ = + − = + − = Rate equations( )
{
(
)
[
]
(
)
[
]
}
( )
{
[
]
[
]
}
( )
{
β[
ατ]
α[
βτ]
}
α β τ τ β τ α α β τ τ β β φ κ τ α α φ κ α β τ − − − − − = − − − − = − − + − − − + − = exp exp 1 1 exp exp 1 exp exp 1 w v u t k k k k k w v u w v u a p w a x v a a u 1 1 2 1 1 0 0 0 0 1 0 1 = = = = = = = = + + ∀ = = = − φ τ κ τ τNote that α and β are composite quantities containing the individual rate constants.
φ κ β α φ αβ + + = + = 1
QSSA assumes that
0
≅
τ
d
dv
(
)
φ κ φ κ + ≅ = + − ss ss ss ss u v v u 0 + − ≅ + − = τ φ κ φ φ κ φ τ exp ss ss ss u u d du φ κ τ φ κ φ + + − ≅ exp ss v + − − = + − + ≅ + − + = ≅∫
τ φ κ φ τ τ φ κ φ φ κ φ τ φ κ φ φ κ φ φ τ τ exp 1 exp exp 0 d w v d dw ss ss ssUsing 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
log τ
0.01 0.1 1 10 100
no
rm
al
is
e
d
c
o
nc
e
n
tr
at
io
n
0.0 0.2 0.4 0.6 0.8 1.0 1.2
u(τ) v(τ) w(τ)
log τ
0.01 0.1 1 10 100
no
rm
al
is
e
d
c
o
nc
e
n
tr
at
io
n
0.0 0.2 0.4 0.6 0.8 1.0 1.2
uss(τ)
vss(τ) wss(τ)
A
X
P
A
X
P
A X P
k1
k-1
k2
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.
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.
log τ
0.01 0.1 1 10 100
no
rm
al
is
e
d
c
o
nc
e
n
tr
at
io
n
0.0 0.2 0.4 0.6 0.8 1.0 1.2
u(τ) v(τ) w(τ)
log τ
0.01 0.1 1 10 100
no
rm
al
is
e
d
c
o
nc
e
n
tr
at
io
n
0.0 0.2 0.4 0.6 0.8 1.0 1.2
uss(τ) vss(τ) wss(τ)
A
P
X A
P
X
A X P
k1
k-1
k2
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.
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.
Macromolecule Formation : Polymerization.
Polymerization of vinyl halides
occurs via a
chain growth addition
polymerization
mechanism.
C H H
C H
X
C H
H C H
X n
3 step process :
•
initiation
•
propagation
•
termination
monomer vinyl halide
X = Cl, vinyl chloride
polymer
poly(vinyl chloride)
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.
Free Radical Addition Polymerization.
Initiation.
reactive freeradicals generated
k
i= initiation rate
constant
k
p= propagation
rate constant
k
T= k
TC+ k
TD= termination rate
constant
⋅
→
+
⋅
⋅
→
12
M
M
R
R
I
kiinitiator
species active monomer
not kinetically significant
⋅
→
+
⋅
⋅
→
+
⋅
+ + + 2 1 1 j k j j k jM
M
M
M
M
M
p pTermination
m n k m n m n k m nP
P
M
M
P
M
M
TD TC+
→
⋅
+
⋅
→
⋅
+
⋅
+ termination via combinationtermination via disproportionation macroradical monomer
Propagation .
Free Radical Addition Polymerization.
Analysis of the steady state kinetics.
Propagating macroradicalconcentration will be smallsince they are very reactive so QSSA can be applied.
polymerization rate = rate of chain propagation
0
≅
dt
dx
removal x
formation x
removal x
formation x
r
r
r
r
dt
dx
=
≅
−
=
0
macroradical concentration
mx
k
r
dt
dm
R
P=
−
=
P=
Pmonomer concentration bimolecular propagation
rate constant
initiation rate ri
T i
r
r
=
2
2
k
x
r
T=
Tri can be left unspecified depends on initiation mechanism
m
k
r
k
mx
k
r
R
T i P
P P
P
2 / 1
2
=
=
=
termination rate rT
2 macroradicals disappear for each
incidence of termination
2
2
k
x
r
i=
TT i
k
r
x
2
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.
∆H
EA lowered
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.
catalyst absent
energy
∆EA
catalyst present
reactants
products
thermodynamics unchanged
O CH2OH
H
OH OH H
H
OH H
H
+ ATP
O
CH2OPO3 2-H
OH OH H
OH
OH H
H H
+ ADP + H+
hexokinase
hexokinase
glucose
hexokinase
glucose
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.
glucose glucose 6- phosphate
binding pocket substrate
Chymotrypsin :
A digestive
enzyme .
active site
Natural
molecular
Michaelis-Menten kinetics.
S + E ES E + P
KM kc
u=c/KM
0 10 20 30 40 50
Ψ =R Σ /k c e Σ 0.0 0.2 0.4 0.6 0.8 1.0 1.2 u u k e R c K c e k R c M c + = = Ψ + = Σ Σ Σ Σ 1 c e K k R u K c u M c M Σ Σ ≅ ≅ Ψ << <<1 1st order kinetics Σ Σ ≅ >> >> ≅ Ψ e k R K c u c M 1 1 zero order kinetics
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).
Ψ
u
Ψ
This is an example of a complex rate equation, where the reaction rate varies with reactant concentration in a non linear way.
Michaelis-Menten rate
equation. ES complex decomposition
to form products substrate
binding enzyme/substrate complex
Experimental rate equation :
R
s
b
as
dt
ds
+
=
−
=
Σ substrate
concentration
a, b = constants Proposed mechanism should produce this rate law.
Mechanism :
E + S
ES
EP
E + P
enzyme
substrate
product enzyme/substrate
complex enzyme/productcomplex
Rate equations
We analyse a simpler scheme which includes all the essentials.
(
)
p
e
k
x
k
dt
dp
p
e
k
s
e
k
x
k
k
dt
dx
x
k
s
e
k
dt
ds
2 2
2 1
1 2
1 1
−
− −
−
−
=
−
−
+
=
−
−
=
−
p = [P] s = [S]
E + S ES E + P
k1
k-1
k2
k-2
What forms does the enzyme take? • free enzyme E
• bound enzyme ES
e
x
e
Σ=
+
Usually eΣ << s . Subsequent to mixing one has an 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 during the course of the reaction.
This requirement satisfies the QSSA.
total initial
enzyme concentration
QSSA
{
1+
1+
2+
2}
−
1−
2≅
0
=
k
s
k
−k
k
−p
x
k
se
Σk
−pe
Σdt
dx
SS(
)
2 1 2 1 2 1k
k
p
k
s
k
e
p
k
s
k
x
SS+
+
+
+
=
− − Σ −{
}
2 1 2 1 2 1 2 1k
k
p
k
s
k
e
p
k
k
s
k
k
dt
ds
R
+
+
+
−
=
−
=
− − Σ − − ΣLet’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.
This has the same form
as the empirical rate
equation observed
experimentally.
0 1 2 1 0 2 0 1 2 1 0 2 1 0 ,s
k
k
k
s
e
k
s
k
k
k
s
e
k
k
R
R
+
+
=
+
+
≅
≅
− Σ − Σ Σ Σinitial rate
Michaelisconstant Fundamental kinetic
parameters : KM and kC .
mol dm-3
0 0 0 ,
s
K
s
e
k
R
M C+
=
Σ Σ 2 1 2 1 k k k k k K C M = + = − catalytic rate constantKM : enzyme/substrate binding
kC : decomposition of
enzyme/substrate complex. Michaelis-Menten (MM)
equation for steady state enzyme kinetics.
s-1
How can we evaluate KM and kC?
• Non linear least squares fitting to MM equation. • Suitable linearization of MM equation.
Σ Σ Σ + = e k s e k K
R C C
M 1 1
1 0 0 , SLB 0 , 1 Σ R Σ
=
e
k
K
S
C M LB 0 1s I = k eΣ
C C M
k
s
k
K
k
1
1
0+
=
Σ unsaturated enzyme kinetics saturated enzyme kinetics 0 0 0 ,s
K
s
e
k
e
k
dt
dp
R
M C+
=
=
=
Σ Σ Σ Σ 0 0s
K
s
k
k
M C+
=
Σ composite rate constantWe 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 US
Case US
0 0 0 0 1 1 1 s K k k s k K k k s k K K s M C C M C C M M ≅ ≅ >> << Σ Σ(
)
θ + = + = + = = − −− 1 1
2 1 1 2 2 1 1 2 1 2
1 K k
k k k k k k k k k K k k M C U 1 2 1 1 1 − − = = k k k k K
θ
equilibrium constantfor ES formation We consider two
1 1
1 2
1 2
1
1 2
1
k K
k k
k k K k
k k
U = =
≅
= ≅ +
− −
−
θ
θ
When k2 >>k-1 , ES complex decomposition to formproducts 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.
θ
>> 1 scenario.
E + S
ES
E + P
k
-1k
1RDS
k
2RDS
k
-1k
2k
1θ
<< 1 scenario.
Have a fast pre-equilibrium ESfollowed by a slow rate determining decomposition of the ES complex to form products.
E + S
E + P
2 1
1
1
k
K
k
U≅
Saturated enzyme kinetics : all active sites
in enzyme bound with substrate.
Have slow rate determining breakdown of
ES complex to form products.
Case S
2 0 0 01
1
1
1
k
k
k
k
k
k
s
k
K
K
s
K
s
C C C C M M M=
≅
≅
<<
>>
>>
Σ ΣThe detailed form of KM and kC in terms of fundamental rate constants will depend on the nature of the mechanism.
E + S
ES
EP
E + P
k
1k
-1k
2k
-2k
3k
-3 C C Mk
s
k
K
R
e
1
1
0 0 ,
+
=
Σ ΣLB inverse plot will always pertain.
Pre equilibrium 3 2 1 2 1 1 1 1 1 k K K k K k k K C
M = + +
Case S
Pre equilibrium
Case US
3 2 3 21
1
1
1
k
K
k
k
k
C=
+
+
adsorbed products
gas phase products
energy
gas phase pathway
EG
gas phase reactants
adsorbed reactants EA
ED
ES
Ej = activation energy for step j.
G : gas phase reaction A : adsorption
S : surface reaction D : desorption
surface catalysed
pathway
Reaction intermediates stabilized via bonding
The Metal-Catalyzed Hydrogenation
of Ethylene
H
2
C
CH
2(g) + H
2(g)
H
3C
CH
3(g)
•
ads
H reaction betweenadsorbed species
ads
CH C
H2 • − 3
•
ads
H
•
ads
H
ads
H C2 4
Adsorption
.
Term used to describe the process whereby a molecule (theadsorbate) forms a bond to a solid surface (an adsorbent).
number of sites occupied by adsorbate
Σ
=
N
N
Sθ
Fractional surface coverage
θ
total number of adsorption sites When θ = 1, NS = NΣ and an
adsorbed monolayer is formed. The fractional coverage θ depends
on pressure of adsorbing gas phase species.
This θ = θ (p) relationship is called an adsorption isotherm.
A (g)
A
adsdynamic equilibrium
surface
Langmuir Adsorption Isotherm.
Langmuir Adsorption Isotherm.
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
associative adsorption
adsorption rate constant A (g) + S Aads
kA
kD
A2 (g) + S 2 Aads kA
kD
desorption rate constant
K measures affinity of a particular
molecule for an adsorption site.
D A
k
k
K
=
dissociative adsorption
surface adsorption site
(
θ
)
Dθ
A
p
k
k
1
−
=
At equilibrium : R
A= R
DRate of adsorption :
Langmuir adsorption isotherm :
associative adsorption.
(
−
θ
)
=
k
p
1
R
A Afractional coverage of vacant sites
pressure
Kp
Kp
+
=
1
θ
Rate of desorption :
θ
D
D
k
R
=
fractional surface coverageKp
1
1
1
=
+
θ
θ
1
1
slope = 1/K
p
A similar analysis can be done for dissociative adsorption.
(
)
2
2
1
θ
θ
D D
A A
k
R
p
k
R
=
−
=
At equilibrium :
(
)
2 21
θ
Dθ
A
D A
k
p
k
R
R
=
−
=
Desorption rate
Adsorption rate
Adsorption isotherm for dissociative
adsorption.
(
)
Kp
Kp
Kp
p
k
k
D A
+
=
=
=
−
1
1
22
θ
θ
θ
Kp
1
1
1
=
+
θ
p
1
KS = 1
θ
P/atm
0.0 0.5 1.0 1.5 2.0
S
ur
fac
e c
o
ve
ra
g
e
θ
0.0 0.2 0.4 0.6 0.8 1.0
K = 10 K = 1 K = 0.1
1
1
1
→
≅
+
>>
θ
Kp
Kp
Kp
high plimit :
monolayer formed
Kp
Kp
Kp
≅
≅
+
<<
θ
1
1
1
Kp
Kp
+
=
1
θ
Langmuir Adsorption Isotherm.
K large
K small
Kinetics of surface reactions.
Assume that gaseous reactant decomposes when it is adsorbed.
θ
k
R
=
Surface coverage of adsorbed gasReaction rate
Surface coverage related to gas pressure p via
Langmuir adsorption isotherm
Kp
Kp
+
=
1
θ
Kp
kKp
R
+
=
1
We can consider two limits.
Low pressures. High pressures.
Rate depends linearly on gas Pressure p
First order kinetics. Adsorption process is rate determining when p is low.
Decomposition is fast.
1
<<
Kp
kKp
R
≅
Rate independent of Gas pressure p
Zero order kinetics.
1
>>
Kp
k
R
≅
Adsorption rate verylarge when p is high.A(g)
Adsorption of a gas on a solid is an exothermic process : ∆ Hads is negative.
Both adsorption and desorption processes follow the Arrhenius equation.
−
=
−
=
RT
E
A
k
RT
E
A
k
D D
D
A A
A
exp
exp
adsorption pre-exponential factor
activation energy for adsorption
E
AE
D∆
H
adsA
adsactivation energy for desorption
D A
ads
ads D
A D
A
E
E
H
RT
H
A
A
k
k
K
−
=
∆
∆
−
=
=
exp
temperature (K)
desorption pre-exponential factor
How is
∆
H
adsmeasured ?
0 0
0
ln
ads adsads
RT
K
H
T
S
G
=
−
=
∆
−
∆
∆
Gibbs energy of adsorption entalphy of adsorption
2 0 0 0
ln
ln
RT
H
T
K
R
S
RT
H
K
ads ads ads∆
=
∂
∂
∆
+
∆
−
=
θ θ θ θ θ θ θ θ θ ∂ ∂ − = ∂ ∂ = ∂ ∂ + ∂ ∂ − = + − = T K T p T p T K p K Kp ln ln 0 ln ln 1 ln ln ln 1Langmuir adsorption assumed
p
0 1 2
θ 0.0 0.5 1.0 − ∆ = ∆ − = ∂ ∂ 2 1 0 2 1 2 0 1 1 ln ln T T R H p p RT H T p ads ads θ θ T2 T1 p1 p2
entropy of adsorption
constant surface coverage
constant surface coverage
p1, p2, T1 and T2 can be measured so
Chemical Kinetics.
Lectures 5-6.
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 E
Ais determined from experiment, by
measuring the rate constant k at a number of different
temperatures. The Arrhenius equation
is used to construct an Arrhenius plot
of ln k versus 1/T. The activation energy
is determined from the slope of this plot.
−
=
RT
E
A
k
exp
APre-exponential factor
( )
=
−
=
dT
k
d
RT
T
d
k
d
R
E
Aln
/
1
ln
2k
ln
T
1
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.
References for Microscopic Theory
of Reaction Rates.
•
Collision Theory
.
– Atkins, de Paula, Physical Chemistry 7
thedition, Chapter 27, Section. 27.1,
pp.944-951.
•
Activated Complex Theory.
– Atkins, de Paula, Physical Chemistry 7
thCollision theory of bimolecular
gas phase reactions.
• We focus attention on gas phase reactions and assume
