C8 REACTION KINETICS

Example :
The half-life of the decomposition of NO2 is 4.0
minutes and it is a second order reaction;
2NO2(g)  2NO(g) + O2(g)

If 7.8 x 10-3 M of NO2 is allowed to decomposed,
calculate :
(a) the rate constant.
(b) time required to reduce the concentration of NO2
to 1.6 x 10 -3 M

Ans : (a) 32.05 M-1 min-1
(b) 15.5 min

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Second order reaction:

1 − 1 = kt
[A] [A]

a) t1Τ2= 1
k[A]

4 = k 1 10−3)
(7.8 ×

k = 32.05 M−1min−1

b) (1.6 1 − (7.8 1 = 32.05 × t
× 10−3) × 10−3)

t = 15.5 min

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Example :
Write rate law for this equation,
A + B  C
i) When [A] is doubled, rate also doubles. But
doubling the [B] has no effect on rate.
ii) When [A] is increased 3x, rate increases 3x,
and increasing of [B] 3x causes the rate to
increase 9x.
iii) Reducing [A] by half has no effect on the rate,
but reducing [B] by half causes the rate to be
half the value of the initial rate.

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ANSWER r1  k Am Bn  equation 1
2r1  k 2Am Bn  equation 2
i r1  k Am 2Bn  equation 3

equation 2 : equation 3 :
equation 1 equation 1

2r1  k 2Am Bn r1  k Am 2Bn
r1 k Am Bn r1 k Am Bn

2  2m 1  2n

m 1 n0

rate  k A1 B0  k A1 54

ANSWER r1  k Am Bn  equation 1
3r1  k 3Am Bn  equation 2
ii 9r1  k Am 3Bn  equation 3

equation 2 : equation 3 :
equation 1 equation 1

3r1  k 3Am Bn 9r1  k Am 3Bn
r1 k Am Bn r1 k Am Bn

3  3m 9  3n

m 1 32  3n

n2

rate  k A B2 55

iii r1  k Am Bn  equation 1
m
 1 
r1  k  2 A  Bn  equation 2

1  1 B n
2  2
r1  k Am  equation 3

equation 2 : equation 3 :
equation 1 equation 1

 1 A m 1  1 B n
 2 2  2
r1 k k Bn r1 k Am
r1
 Am Bn r1  k Am Bn

1   1 m Rate  k B1 1   1 n
2 2 2

m0 n 1 56

Determining the Order of Reaction
The order of reaction can be determined by using :

(a) initial rate method
(b) half life based on the graph of concentration

against time.
(c) linear graph method based on the integrated

rate equation and the rate law

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(a) initial rate method

– is used when the concentration of reactants
and their initial rates data are given in a table;
in sets.

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Example :

The following data were measured for the reaction
of nitric oxide with hydrogen:

2NO(g) + 2H2(g)  N2(g) + 2H2O(g)

Number Of Concentration(M) Initial Rate
Experiment (Ms-1)
NO H2
1 0.10 0.10 1.23 x 10-3

2 0.10 0.20 2.46 x 10-3

3 0.20 0.10 4.92 x 10-3

i. Determine the rate law for this reaction.

ii. Calculate the rate constant.

iii. Calculate the rate when [NO] = 0.05 M and

[H2] = 0.150 M 59

i) Determine the rate law for this reaction.
rate = k [NO] [H2]
X = order respect to NO
Y = order respect to H2

Exp 3 ÷ Exp 1

rate 1 = k[NO] [H2]
rate 3 k[NO] [H2]y

4.92 × 10−3 = k (0.20) (0.10)
1.23 × 10−3 k (0.10) (0.10)

4 = (2)

x =2

Second order respect to NO 60

y=1
First order respect to H2

Rate = k [NO]2[H2]

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ii) Calculate the rate constant.

rate = k [NO]2 [H2]

k= rate
[NO]2 [H2]

= 1.23 × 10−3
(0.10)2(0.10)

= 1.23 M−2s−1

iii) Calculate the rate when [NO] = 0.05 M and

[H2] = 0.150 M
rate = k [NO]2 [H2]
= 4.61× 10−4 M s−1

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(b) The half life method (graph [ ] vs t )
The order of reaction can be found from the plot of
the concentration of the reactant against time.
(i) Zero-order reaction

[A]

[A]o

½ [A]o

¼[A]o

t½ (I) t½ (I) time,t 63

(ii) First-order reaction

[A]
[A]o

½[A]o

¼[A]o

t½ (I) t½ (II) time,t

First order-reaction : 64
First half life = Second half life
Thus, t½ (I) = t½ (II)

(iii) Second-order reaction

[A]

[A]o

½[A]o
¼[A]o

t½ (I) t½ (II) time,t

Second-order reaction : 65
t½ (II) = 2 t½ (I)

Example :

When ammonium cyanate is heated strongly, it
decomposes into urea.

NH4NCO  (NH2)2CO

The following results were obtained.

Time(min) 0 20 50 80 150
0.27 0.18 0.13 0.09
[NH4NCO] (M) 0.38

Plot a suitable graph to determine:
(i) The order of reaction
(ii) The half life

66

a) It is a 2nd order reaction
due to t½ (2)=2 t½ (1)

b) t1/2 = 45 min

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(c) linear graph method based on the integrated
rate equation and the rate law

Linear graph method based on the integrated rate equation

• Plot an appropriate graph of the integrated rate
equations

Reaction Intergrated rate Plot graph
order equation
Zero [A] vs t
First [A]o – [A] = kt ln [A] vs t
ln [A]o – ln [A] = kt 1 vs t
Second
1  1  kt A
[ A] [ A]o

• A linear graph indicates the most suitable

order of reaction. 68

Example :
The data below refers to the decomposition of a substance X.

Time (s) 0 100 200 300 400 600
[X]/M 0.1 0.075 0.0558 0.0445 0.0378 0.0297
Plot a suitable graph and determine the order of reaction.

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Second t1 is double from first t1
22

 Second order reaction

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Linear graph method based on the rate law
• Plot a graph rate vs [ ]

(i) Zero order (ii) First order

Rate Rate

[A] [A]

(iii) Second order

Rate Rate

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[A] [A]2

Example :

The data below refers to the decomposition of

hydrogen peroxide.
2H2O2  2H2O + O2

[H2O2]/M 0.1 0.2 0.4 0.6

Rate/ 3.64 x10-5 7.41 x10-5 14.6 x10-5 22.1 x10-5
Ms-1

Plot a graph of rate against concentration and
use your graph to determine :
(a) the order with respect to H2O2
(b) the rate constant

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Example :

The data below refers to the decomposition of

hydrogen peroxide.
2H2O2  2H2O + O2

[H2O2]/M 0.1 0.2 0.4 0.6

Rate/ 3.64 x10-5 7.41 x10-5 14.6 x10-5 22.1 x10-5
Ms-1

Plot a graph a suitable graph to determine :

(a) the order with respect to H2O2
(b) the rate constant

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a) Based on the graph rate
vs [H2O2], it is a first
order reaction. For a
first order reaction,
rate = k [H2O2],
Thus the order with
respect to H2O2 is 1.

b) k= 3.7 x 10-4 s-1

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8.2 Collision Theory

Learning Outcomes:

(a) Explain collision theory
(b) Define activation energy
(c) Explain transition state theory
(d) Draw energy profile diagram of a reaction

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The explanation of the reaction and its rate are
based on :

1. The collision theory
2. The transition theory

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1. Collision Theory

• In a chemical reaction molecules of reactants
collide with each other to form product.

• However not all collisions able to form product.
This can be explain by the Collision Theory

• The Collision theory is based on the ideas that:

(i) Molecules of reactants must collide in
order to form products.

(ii) Products are formed only when effective
collisions between molecules occur.

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Effective collisions

Not all collisions between the reactant molecules
are effective collisions. A collision is an effective
collision when

i. The colliding molecules have
a total kinetic energy equal to or
greater than the activation energy, Ea.

ii. The molecules must collide in the
correct orientation.

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(i) Activation Energy

• Activation energy, Ea is the minimum energy
required for effective collisions to occur to initiate
a chemical reaction.

• The activation energy, Ea may be gained
through the collisions of molecules and is used
to break the bonds between atoms of the
reactants.

• Then, these atoms will form new bonds in the
product.

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(ii) The orientation of collisions

• The correct orientation means that the collision
of molecules occur at the correct angle/manner
that favors the formation of products.

+

+

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2. Transition State Theory

• Explains the activation energy, Ea as the energy
required by the reactant molecules to form an
activated complex.

• An activated complex is a temporary species
formed by the reactant molecules as a result of
the collision before they form the products.

• The transition theory can be explained using the
energy profile diagram

81

Exothermic Reaction

Energy Activated
complex

Ea

Reactants

∆H= -ve

Products

Progress of reaction

82

Endothermic Reaction

Energy Activated
complex

Ea
Products

∆H= +ve

Reactants

Progress of reaction

83

• The activated complex or transition state ( an
intermediate stage that lies between the reactant
and the product) formed temporarily in an
effective collision when reactants molecules
achieve the activation energy,Ea

• It is unstable high energy species can either
form products or re-form reactants in which
old bonds of the reactants are partially broken
and new bonds of the products are partially
formed.

84

• Example:
H2 (g) + I2 (g)  2HI (g) ∆H=+ve

HI ⇌ HI HI

+ ⇌ +
HI HI
HI

Activated
complex

85

Endothermic Reaction

HI

HI Bond breaking
Bond forming
Energy

Ea 2HI(g)
H2(g) + I2(g)
∆H= +ve

Progress of reaction 86

Collision Theory and Transition State Theory

87

8.3 Factors affecting reaction rate

Learning Outcomes:

(a) Explain the effect of the following factors on the
reaction rate:

i. concentration or pressure
ii. temperature

iii. catalyst
iv. particle size
(b) Illustrate the effect of temperature on reaction
rate using Maxwell-Boltzmann distribution
curve.
(c) Compare the curve of the energy profile
diagram for a reaction with and without catalyst.

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Learning Outcomes:

- Ea

(d) State Arrhenius equation k  Ae RT

(d) Explain the relationship between temperature
and activation energy to the rate constant
based on the Arrhenius equation.

(d) Determine k, Ea , T and A using Arrhenius
equation by calculation and graph.

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FACTORS THAT AFFECT THE
REACTION RATE

1. Concentration/Pressure

Concentration :
• Reaction rates generally increase as the

concentrations of the reactants are increased.
• Concentration of reactants increases
• number of molecules per unit volume

increases
• number of effective collisions increases
• reaction rate increases

90

Pressure (for gaseous reactants) :
• Pressure increases, volume decreases
• number of molecules per unit volume increases,

number of effective collisions increases,
• reaction rate increases

91

2. Temperature

 The effect of temperature on the reaction rate
can be explained in terms of kinetic theory.

• The average kinetic energy of molecules are
proportional to the temperature.

• At higher temperature, more molecules will have
higher kinetic energies than at lower
temperature.

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At higher temperature :

• The number of molecules having kinetic energy
equal to or more than activation energy, Ea will
increases

• The frequency of effective collision increases
• Reaction rate increases
• Maxwell-Boltzmann distribution shows the

kinetic energy distributions for a reaction
mixture at two different temperature.

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Maxwell-Boltzmann Distribution of Kinetic Energies

No. of Lower temperature, T1
molecules

Higher temperature, T2

Molecules with enough
kinetic energy to react

Ea Kinetic energy

Number of molecules with energy ≥ Ea at T2 94
Number of molecules with energy ≥ Ea at T1

• The shaded areas represent the number of
molecules having kinetic energy equal to or
greater than the activation energy, Ea.

• At the higher temperature,The number of
molecules having kinetic energy equal to or
more than activation energy, Ea increases

• Number of collision per unit time increases
• rate of reaction increases.

95

3. Catalyst
• A catalyst is a substance that increases the rate

of a chemical reaction without itself being
consumed.

• A catalyst provides an alternative pathway which
has a lower activation energy compared to the
one without catalyst

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Comparison of the activation energy of the
catalyzed and uncatalyzed reactions

Energy E 1 = Ea without catalyst
E 2 = Ea with catalyst

E1

Reactants E2

products

Reaction Progress

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4. Particle Size
• Size of reacting particles decreases,
• total surface area exposed for reaction
increases
• the faster the reaction.
• reaction rate increases

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THE ARRHENIUS EQUATION

• The relationship between the rate constant of a
reaction and temperature can be expressed by the
Arrhenius equation.

- Ea

k  Ae RT

Ea = activation energy 99
k = rate constant
R = gas constant, 8.314 J mol-1K-1

T = absolute temperature, K

e = the base of the natural logarithm scale

A = quantity represents the collision frequency
(frequency factor)

• This equation can be expressed in a more useful
form by taking the natural logarithm of both sides.

- Ea

k  Ae RT

ln k  ln A - Ea
RT

• The equation shows that:
(a) the rate constant increases as temperature(T)
increases
(b) the rate constant decreases as activation
energy(Ea) increase

100