1
Chapter 12
1
Chemical Kinetics – chapter outline
1. Introduction on chemical kinetics
2. Rate of Reaction
- definition
- type of reaction rate
- the differential rate equation
3. Rate constant
4. Rate Law
- rules of rate law
5. Order of Reaction
- zero order
- first order
- second order
6. Reaction Mechanisme
- molecularity
- rate law and elementary step
- rate determining step
7. Collision Theory
- effective collision
- activation energy
- transition state theory
- characteristic of activated
complex
8. Factor Affecting Reaction Rates
- temperature
- cooncentration
- catalyst
- surface area
- pressure
2
Chemical Kinetics – chapter outline
8. Nuclear Chemistry
- the discovery of radiation
- types of radiation
- penetrating power of radiotion
- type of nuclear reaction
- nuclides
- elementary particle
- radioactive decay
- nuclear transmutation
9. Half life
- half-life in radioactive decay
- half-life in chemical reaction
3
Introduction on
Chemical Kinetics
4
Definition
Chemical kinetics is the study of the rates of chemical reactions,
the factors that affect these rates, and the reaction mechanisms
by which reactions occur.
Thermodynamics – does a reaction take place?
Kinetics
– how fast does a reaction proceed?
5
Rate of Reaction
6
Definition of Reaction Rate
 Reaction rate is the change in the concentration of a reactant or a
product with time
 change in concentration/ change in time
 Unit: M/s
reactant
(with respect to A)
(with respect to B)
d[A]
=dt
= +d[B]
dt
A
B
product
(Rate of disappearance of A)
(Rate of appearance of B)
 d[A] is negative, because [A] decreases w
d[B] is positive, because [B] increase with
7
4
A
B
Time ↑
d[A]
rate = - dt
rate =
d[B]
dt
8
Type of Reaction Rate
1. Average rate is the rate over a period of time.
2. Instantaneous rate is the rate of reaction at a given time
Instantaneous rate is determined from a graph of concentration
vs time by drawing a line tangent to the curve at that particular
time.
3. Initial rate is the instantaneous rate at the beginning of a
reaction
9
Reaction:
H2O2(aq)  H2O(l) + ½ O2(g)
 Reaction rates are obtained from
the slopes of the straight lines
purple
An average rate from the
purple line.
The instantaneous rate at t
=300 s from the red line.
 The initial rate from the blue
line.
10
The Differential Rate Equation
 A differential equation enables the relationship between the
rate of disappearance of reactants and the formation of
product
 Relation between reaction rate and stoichiometry of reaction
11
Example:
2A
1 d[A]
rate = 2 dt
In General:
aA + bB
B
d[B]
rate = dt
cC + dD
1 d[A] = - 1 d[B]
1 d[C] = 1 d[D]
rate = =
dt
dt
a
b
c dt
d dt
* a,b,c and d are the stoichiometric
coefficients
12
Exercise 1
a)
b)
13
Exercise 2
Consider the reaction: 2 HI
H2 + I2.
Determine the rate of disappearance of HI when the rate of I2
formation is 1.8 x 10-6 M s-1.
Answer: 3.6 x 10-6 M14 s-1
Exercise 3
Consider the reaction: 2H2 + O2
2H2O
a. Express the rate in terms of changes in [H2], [O2] and
[H2O] with time.
b. When [O2] is decreasing at 0.23 M s-1, at what rate is
[H2O] increasing?
Answer: 0.46 M15s-1
Rate Constant (k)
16
 Rate Constant (k) is a constant of proportionality between the
reaction rate and the concentration of reactant
 The unit of rate constant is depending on the order of reaction
 Example: Reaction between molecular bromine and formic acid
Br2 (aq) + HCOOH (aq)
brown
2Br - (aq) + 2H+ + CO2 (g)
colourless
17
k value is similar ≈ constant
rate  [Br2]
rate = k [Br2]
rate
Rate constant, k =
[Br2]
= 3.50 x 10-3 s-1
18
Rate Law
The Rate Law expresses the relationship of the rate of
a reaction to the rate constant and the concentration of
the reactants raised to some powers
19
Example:
aA + bB
cC + dD
Rate = k [A]x[B]y
k is rate constant
13
20
Rate = k [A] x[B] y
F2 (g) + 2ClO2 (g)
2FClO2 (g)
Rate = k [F2][ClO2]
Rate Law
Rules
1
21
Example: F2(g) + 2ClO2 (g)
2ClO2 (g)
Data for the reaction between F2 and ClO2
Expt
[F2], M
[ClO2], M
Initial Rate (Ms-1)
Compare Exp .1 & Expt. 2
1
0.10
0.010
1.2 x 10-3
2
0.10
0.040
4.8 x 10-3
Expt. 1 = 1.2 x 10-3 = (0.1)x (0.01)y
Expt. 2
4.8 x 10-3 (0.1)x (0.04)y
3
0.20
0.010
2.4 x 10-3
In general, Rate = k [F2]x [ClO2]y
Compare Exp .1 & Expt. 3
0.25 = (0.01)y
(0.04)y
0.25 = (0.25)y
Hence, y = 1
Expt. 1 = 1.2 x 10-3 = (0.1)x (0.01)y
Expt. 3
2.4 x 10-3 (0.2)x (0.01)y
0.5 = (0.1)x
(0.2)x
0.5 = (0.5)x
* Insert x & y value into the general eq.
Rate = k [F2]1[ClO2]1
Hence, x = 1
22
Exercise 4
Answer: k= 0.08 M-123s-1
Order of Reaction
1. Zero order
2. First order
3. Second order
24
Zero Order
[ ]
25
- d[A] = k
dt ��
-[A] = kt – [A]o
Hence
[A]o – [A]t = kt
[A]o is the concentration of A at time t = 0
[A] is the concentration of A at any given time
26
Zero-Order Reactions
[A]o
Rate
[A]
k
[A]
k
Time
27
First Order
[ ]
28
ln [A]o -ln [A]t = k t
Hence
ln [A] = k t + c
t
29
First -Order Reactions
[A]
t
ln[A]
ln[A]o
[A]
ln[A]o
t
t
30
Second Order
[ ]
1 - 1 = kt
[A]
[A]o
31
Second -Order Reactions
[A]
t
1/[A] – 1/[A]o
1/[A]
1/[A]o
t
t
32
Exercise 5:
Determination of the orders of reaction;
O2(g) + 2NO(g)
2NO2(g)
Initial Reactant
Concentrations (mol L-1)
Experiment
O2
NO
Initial Rate
(M s-1)
1
1.10x10-2
1.30x10-2
3.20x10-3
2
2.20x10-2
1.30x10-2
6.40x10-3
3
1.10x10-2
2.60x10-2
12.8x10-3
4
3.30x10-2
1.30x10-2
9.60x10-3
5
1.10x10-2
3.90x10-2
28.8x10-3
Answer: order of reaction =3
33
34
Exercise 6
The reaction 2A
B is first order in A with a rate constant of
2.8 x 10-2 s-1 at 800C. How long will it take for A to decrease
from 0.88 M to 0.14 M ?
Answer: t= 65.65 s
35
Reaction Mechanisme
36
Reaction
Mechanisms
An intermediate is always
formed in an early elementary
step and consumed in a later
elementary step.
Intermediates are species that
appear in a reaction mechanism
but not in the overall balanced
equation.
The overall progress
of a chemical reaction
can be represented at
the molecular level by
a series of simple
elementary steps or
elementary reactions
The sequence of
elementary steps that
leads to product
formation called
reaction mechanism.
37
Example:
Elementary step:
NO + NO
N2O2
+ Elementary step:
N2O2 + O2
2NO2
2NO + O2
2NO2
Overall reaction:
N2O2 (intermediates) is detected during the
reaction but doesn’t exist at the end of reaction
38
Molecularity
 The number of molecules reacting in elementary step
 3 type of molecularity
1. Unimolecular reaction – elementary step with 1 molecule
2. Bimolecular reaction – elementary step with 2 molecules
3. Termolecular reaction – elementary step with 3 molecules
39
Rate Law and Elementary Steps
Rate (elementary step): power raised by [reactant] is depending
on stoichiometric coefficient
Rate (overall reaction): power raised by [reactant] is depending
on experimental value
40
Rate Determining Step
41
Example
Consider the following reaction: 2NO(g) + Cl2(g) → 2NOCl.
The rate law for the above reaction has been determine to be. Rate = k [NO][Cl2]
A mechanisme involving the following steps has been proposed for the above
reaction
Step 1:
NO (g) + Cl2 (g)
Step 2:
NOCl2 (g) + NO (g)
k1
k2
NOCl2 (g)
2 NOCl (g)
a) Write the equation for the overall reaction.
b) Identify the intermediates.
c) Based on the rate law given, which step is the rate determining step
d) What can you say about the relative rates of steps of step 1 and 2.
42
Answer
Consider the following reaction: 2NO(g) + Cl2(g) → 2NOCl.
The rate law for the above reaction has been determine to be. Rate = k [NO][Cl2]
A mechanisme involving the following steps has been proposed for the above
reaction
a)
Write the equation for the overall reaction.
Step 1:
NO (g) + Cl2 (g)
k1
NOCl2 (g)
Step 2:
NOCl2 (g) + NO (g)
k2
2 NOCl (g)
Overall: 2 NO (g) + Cl2 (g)
2 NOCl (g)
b) Identify the intermediates.
NOCl2
** Intermediates are species that appear in a
reaction mechanism but not in the overall
balanced equation.
43
Answer
Consider the following reaction: 2NO(g) + Cl2(g) → 2NOCl.
The rate law for the above reaction has been determine to be. Rate = k [NO][Cl2]
A mechanisme involving the following steps has been proposed for the above
reaction
c) Based on the rate law given, which step is the rate determining step
Step 1:
NO (g) + Cl2 (g)
Step 2:
NOCl2 (g) + NO (g)
k1
k2
NOCl2 (g)
2 NOCl (g)
Rate = k1[NO][Cl2]
Rate = k2[NOCl2] [NO]
• Rate for Step 1 is the same with Experimental Rate (given).
Therefore Step 1 is the ‘rate determining step’
d) What can you say about the relative rates of steps of step 1 and 2.
k1 <<< k2 because’ rate determining step’ is the slowest step in mechanisme
sequence
44
Exercise 7:
Step 2:
k1
slow
k2
fast
a) Write the equation for the overall reaction.
b) Identify the intermediates.
c) Based on the rate law given, which step is the rate determining step
d) What can you say about the relative rates of steps of step 1 and 2.
45
46
Exercise 8:
A proposed mechanism for a reaction is
C4H9Br
→
C4H9+ + H2O
→
C4H9OH2+ + H2O →
C4H9+ + BrC4H9OH2+
C4H9OH + H3O+
slow
fast
fast
a) Write the rate law expected for this mechanism.
b) Write the overall balanced equation for the reaction?
c) What are the intermediates in the proposed mechanism?
47
48
Collision Theory
49
50
Effective Collision
51
Importance of Orientation & Energy
Two molecule that collide must be oriented correctly so that, chemical
change will occured
Example:
52
Importance of Orientation
53
Activation Energy, Ea
 Activation energy is the
minimum energy required to
initiate the chemical reaction
 Activation energy is the
minimum energy required to
performed effective collision
The kinetic energy impact
can be used to stretch, bend
and break covalent bonds
 Molecules that are moving
too slowly collide with
insufficient energy
54
Transition
State Theory
The configuration of the
atoms of the colliding
species at the time of the
collision is called the
transition state.
Species formed at
transition state is
called activated
complex.
Transition
State
Activated
Complex
55
Characteristic of Activated Complex
Very unstable.
It has a short half-life.
Its potential energy is greater than reactants or products.
The activated complex and the reactants are in chemical
equilibrium.
It decomposes to form products or reactants.
56
Potential energy
A reaction profile shows potential
energy plotted as a function of the
progress of the reaction.
Activated Complex
The difference in potential energies
between the products and the
reactants is -∆H for the reaction.
Ea
Reactant
- ∆H
Reactant molecules must have
enough energy to overcome an
energy barrier separating products
from reactants, Ea.
Product
Progress of reaction
57
Factors Affecting
Reaction Rates
1. Temperature
2. Concentration
3. Catalyst
4. Surface Area
5. Pressure
58
Effect of Temperature
Temperature increases
The average kinetic energy of particles increases
Frequency of collision increases
No of particles with kinetic energy more than
activation energy increases
Number of effective collisions increases
Rate of reaction increases
59
Effect of Temperature – Maxwell Boltzman Curve
T1
Where T2 > T1
T2
When T ↑,
Mole frac on ↑ at Ea
Energy increase
Ea
60
Effect of Temperature – Arrhenius Equation
In 1889, Svante Arrhenius proposed the following mathematical
expression for the effect of temperature on the rate constant, k:
where…
k = rate constant
A = constant known as the collision frequency
factor
e = natural log exponent
Ea = activation energy for the reaction
R = universal gas constant (8.314 J K-1 mol-1 )
T = absolute temperature
61
The relation ship between the rate constant, k and
temperature can be seen in the k vs T graph:
1/T (K-1)
62
63
Graph Representation Of The Arrhenius Equation
 Ea 1
ln k 
( )  ln A
R T
Where,
Ea = Activation Energy
R = 8.314 Jmol-1K-1
T = Absolute Temp
A = Collision freq. factor
64
 From previous discussion , we can conclude if the reaction are performed at
specific temperature we can use the equation below
 Ea 1
ln k 
R T
- ln A
 If the value of A (collision frequency factor) is known and the same reaction conducted
at two different temperatures. The Arrhenius equation at each temperature can be
written and combined to formed the equation shown in the box
k
ln
k
1
2
E

a
1
R T
1

2
T
1
65
Exercise 9
Consider the following reaction :
2 HI (g)  H2(g) + I2(g)
The decomposition of hydrogen iodide has rate constants of 9.51 x 10-9 L
mol-1 s-1 at 500 K and 1.10x10-5 L mol-1 s-1 at 600 K. Find Ea.
Answer: 176 kJ mol-1
66
Effect of Concentration
Concentration increases
Number of reacting molecules increases
Frequency of collision increases
Number of effective collisions increases
Rate of reaction increases
67
The frequency of collision increases when the concentration
increases
4 particle system
5 particle system
(2 and 2) → 4 collision
(3 and 2) → 6 collision
68
Effect of Catalyst
A catalyst is a substance that increases the rate of a
chemical reaction without itself being consumed.
A catalyst provides an alternative pathway for the reaction.
This alternative pathway has a lower activation energy.
More colliding particles will have energy equal or greater
than the activation energy.
Rate of effective collision increases.
Rate of reaction increases.
69
uncatalyzed
catalyzed
70
Effect of Surface Area
 The smaller the size of reacting particles, the greater is the
total surface area exposed for the reaction.
 The greater the number of collisions.
 The rate of reaction is the faster.
Small particle, greater surface area
71
Effect of Pressure
Pressure increases
The volume occupied by the gas molecules decrease
The molecules are brought closer together
Frequency of collisions increases
No of effective collisions increases
Rate of reaction increases
Reactant particles far apart
Reactant particles closer
together
72
Nuclear Chemistry
73
The Discovery of Radiation
In 1896 Henri Becquerel made an important discovery.
His work with uranium salts lead to the conclusion that
the minerals gave off some sort of radiation.
74
Following Becquerel’s discovery, in 1898 Marie Sklodowska
Curie and her husband, Pierre Curie, attempted to isolate the
“radioactive” material from the uranium ore.
They discovered two new elements, Radium and Polonium,
both of which were more radioactive than the original ore.
75
In 1911, Rutherford investigated this new property of matter
and discovered that, in the process of emitting radiation,
atoms of one element became atoms of another element.
76
Types of Radiation
77
i. Properties of the different types of radiation
The differences between the three types of radiation can
be seen by passing them through an electric field.
78
ii. Characteristics of an alpha particle
The alpha particle is
deflected to some extent
toward the negative plate.
This indicated that it is
positively (+) charged and has
a fairly large mass.
α particle is same as the
nucleus of a He atom.
two protons
two neutrons
79
iii. Characteristics of an beta particle
The beta particle (b) is deflected
toward the positive plate (+). It is
also deflected more than the a
particle.
This indicates that it is negatively (-)
charged and has a much smaller
mass than the α particle.
β particle is the same as an
electron.
80
iii. Characteristics of an gamma particle
Gamma radiation is not deflected
toward either the positive (+) or
negative (-) plate.
This indicates that it has no
charge.
Gamma rays are a type of
electromagnetic radiation made
up of photons (packets of energy).
81
Penetrating Power of Radiation
The ability of radioactive particles to pass through air and other materials is
inversely related to their mass.
Alpha particles – the least penetrating, they travel only a few centimeters
through air. They can be stopped by a single sheet of paper.
Beta particles – more penetrating, they travel several meters through air.
They can be stopped by a sheet of Al or plastic.
Gamma Rays – most penetrating, thick sheets of lead or concrete are
required to stop gamma rays.
82
Nuclides
Nuclides  a particular type of nucleus, characterized by a
specific atomic number and nucleon number
Nucleon number or mass number  the number of nucleons
(protons and neutrons) in the nucleus of a nuclide.
83
Type of Nuclear Reaction
1. Radioactive decay
Radioactive decay is the process in which an unstable atomic nucleus loses
energy by emitting ionizing/elementary particles (known as radiation),
transforming the parent nuclide atom into a different atom called daughter
nuclide.
2. Nuclear transmutation
Another type of radioactivity results from the bombardment of nuclide by
elementary particles turns into a different atom.
All elements having an atomic number greater than 83 are radioactive !!
84
Elementary Particles
 Also known as ionizing particle or emitting particle
Name
alpha particle
beta particle
(electron)
Notation
4 He
2
0
-1
e
gamma
radiation
proton
positron
or
4α
2
or
0
-1
0
0
neutron
Symbol
β
0
+1
e
or
or
βγ
γ
n
1n
0
1H
1
α
1p
1
0
+1
β
p
β+
85
Radioactive Decay
1. Alpha() Decay
 Involve the lost of α particle,
4 He
2
from a nucleus
 For every α particle emitted by the parent nucleus
i) A (atomic number) will decreases by 4
ii) Z (mass number) will decreases by 2
 Example
+
proton
neutron
Uranium
Thorium
86
2. Beta(- ) Decay
 Involve the lost of β- particle,
0
-1
e
from a nucleus
 For every β- particle emitted by the parent nucleus
i) A (atomic number) will increases by 1
ii) Z (mass number) remains
 Example
+
proton
neutron
Thorium
Protactinium
 How does nucleus emit an ē?
- neutron changes into proton & emits electron
0
-1
e
87
3. Positron (+ ) Decay
 Involve the lost of β+ particle,
0
+1
e
from a nucleus
 For every β+ particle emitted by the parent nucleus
i) A (atomic number) will decreases by 1
ii) Z (mass number) remains
 Example
95 Tc
43
95 Mo
42
+
0β
1
 How does nucleus emit an ē?
- proton changes into neutron & emits positron
88
4. Gamma (γ) emission
 Involve the lost of γ particle, from a nucleus
 For every γ particle emitted by the parent nucleus
i) A (atomic number) remains
ii) Z (mass number) remains
 Example
40
18
Ar → 1840Ar + 00 γ
• Electromagnetic radiation
• Very high-energy photons
• It has no charge & no mass
 Gamma radiation does NOT change number of proton or neutron
in a nuclide but it does change the way they are arranged
89
Nuclear Transmutation
1. Electron capture
 Conversion of proton to neutron by electron bombardment
 For every electron particle bombardment on the parent nucleus
i) A (atomic number) will decreases by 1
ii) Z (mass number) remains
 Example : Bombardment of Beryllium-7 with electron
7
4Be
+
0
-1e

7
3Li
90
2. Bombardment of neutron particle
 Example:
6
3Li
1
3
4
+ 0n  1H + 2α
3. Bombardment of alpha particle
 Example:
14
7N
+
4
2α
17
 8o
1
+ 1p
91
Half Life
92
Half-life (t1/2 ) in Radioactive Decay
The time for half amount of the radioactive nuclei in a given
sample to undergo decay.
For any radioisotope:
No. of half-life
% Remaining
0
100
1
50
2
25
3
12.5
4
6.25
5
3.125
6
1.5625
93
Example:
94
Half-life (t1/2 ) in Chemical Reactions
The half-life, t½ in chemical reaction is the time required for the
concentration of a reactant to decrease to half of its initial
concentration.
Order of
Reaction
0
Rate Law
rate = k
1
rate = k [A]
2
[A]2
rate = k
Concentration-Time
Equation
[A] = [A]0 - kt
ln[A] = ln[A]0 - kt
1
1
=
+ kt
[A]
[A]0
Half-Life
t½ =
[A]0
2k
t½ = ln2
k
1
t½ =
k[A]0
95
Exercise 10
What is the half-life of N2O5 if it decomposes with a rate constant of
5.7 x 10-4 s-1?
Answer: t ½ = 1216 s
96