Chemistry, by its very nature, is concerned with change.
Substances with well defined properties are converted
by chemical reactions into other substances with
different properties. For any chemical reaction, chemists
try to find out
(a) the feasibility of a chemical reaction which can be
predicted by thermodynamics ( as you know that a
reaction with ∆G < 0, at constant temperature and
pressure is feasible);
(b) extent to which a reaction will proceed can be
determined from chemical equilibrium;
(c) speed of a reaction i.e. time taken by a reaction to
reach equilibrium.
Along with feasibility and extent, it is equally
important to know the rate and the factors controlling
the rate of a chemical reaction for its complete
understanding. For example, which parameters
determine as to how rapidly food gets spoiled? How
to design a rapidly setting material for dental filling?
Or what controls the rate at which fuel burns in an
auto engine? All these questions can be answered by
the branch of chemistry, which deals with the study
of reaction rates and their mechanisms, called
chemical kinetics. The word kinetics is derived from
the Greek word ‘kinesis’ meaning movement.
Thermodynamics tells only about the feasibility of a
reaction whereas chemical kinetics tells about the rate
of a reaction. For example, thermodynamic data
indicate that diamond shall convert to graphite but
in reality the conversion rate is so slow that the change
is not perceptible at all. Therefore, most people think
After studying this Unit, you will be
able to
• define the average and
instantaneous rate of a reaction;
• express the rate of a reaction in
terms of change in concentration
of either of the reactants or
products with time;
• distinguish between elementary
and complex reactions;
• differentiate between the
molecularity and order of a
reaction;
• define rate constant;
• discuss the dependence of rate of
reactions on concentration,
temperature and catalyst;
• derive integrated rate equations
for the zero and first order
reactions;
• determine the rate constants for
zeroth and first order reactions;
• describe collision theory.
Objectives
Chemical Kinetics helps us to understand how chemical reactions
occur.
4
Chemical Kinetics
Unit
Unit
Unit
Unit
Unit
4
Chemical Kinetics
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96Chemistry
that diamond is forever. Kinetic studies not only help us to determine
the speed or rate of a chemical reaction but also describe the
conditions by which the reaction rates can be altered. The factors
such as concentration, temperature, pressure and catalyst affect the
rate of a reaction. At the macroscopic level, we are interested in
amounts reacted or formed and the rates of their consumption or
formation. At the molecular level, the reaction mechanisms involving
orientation and energy of molecules undergoing collisions,
are discussed.
In this Unit, we shall be dealing with average and instantaneous
rate of reaction and the factors affecting these. Some elementary
ideas about the collision theory of reaction rates are also given.
However, in order to understand all these, let us first learn about the
reaction rate.
Some reactions such as ionic reactions occur very fast, for example,
precipitation of silver chloride occurs instantaneously by mixing of
aqueous solutions of silver nitrate and sodium chloride. On the other
hand, some reactions are very slow, for example, rusting of iron in
the presence of air and moisture. Also there are reactions like inversion
of cane sugar and hydrolysis of starch, which proceed with a moderate
speed. Can you think of more examples from each category?
You must be knowing that speed of an automobile is expressed in
terms of change in the position or distance covered by it in a certain
period of time. Similarly, the speed of a reaction or the rate of a
reaction can be defined as the change in concentration of a reactant
or product in unit time. To be more specific, it can be expressed in
terms of:
(i) the rate of decrease in concentration of any one of the
reactants, or
(ii) the rate of increase in concentration of any one of the products.
Consider a hypothetical reaction, assuming that the volume of the
system remains constant.
R → P
One mole of the reactant R produces one mole of the product P. If
[R]
1
and [P]
1
are the concentrations of R and P respectively at time t
1
and [R]
2
and [P]
2
are their concentrations at time t
2
then,
∆t = t
2
– t
1
∆[R] = [R]
2
– [R]
1
∆ [P] = [P]
2
– [P]
1
The square brackets in the above expressions are used to express
molar concentration.
Rate of disappearance of R
[
]
Decrease in concentration of R
R
=
Time taken
t
∆
= −
∆
(4.1)
4.14.1
4.14.1
4.1
Rate of aRate of a
Rate of aRate of a
Rate of a
ChemicalChemical
ChemicalChemical
Chemical
ReactionReaction
ReactionReaction
Reaction
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Chemical Kinetics
Rate of appearance of P
[
]
Increase in concentration of P
P
=
Time taken
t
∆
= +
∆
(4.2)
Since, ∆[R] is a negative quantity (as concentration of reactants is
decreasing), it is multiplied with –1 to make the rate of the reaction a
positive quantity.
Equations (4.1) and (4.2) given above represent the average rate of
a reaction, r
av
.
Average rate depends upon the change in concentration of reactants
or products and the time taken for that change to occur (Fig. 4.1).
Fig. 4.1: Instantaneous and average rate of a reaction
Units of rate of a reaction
From equations (4.1) and (4.2), it is clear that units of rate are
concentration time
–1
. For example, if concentration is in mol L
–1
and
time is in seconds then the units will be mol L
-1
s
–1
. However, in gaseous
reactions, when the concentration of gases is expressed in terms of their
partial pressures, then the units of the rate equation will be atm s
–1
.
From the concentrations of C
4
H
9
Cl (butyl chloride) at different times given
below, calculate the average rate of the reaction:
C
4
H
9
Cl + H
2
O → C
4
H
9
OH + HCl
during different intervals of time.
t/s 0 50 100 150 200 300 400 700 800
[C
4
H
9
Cl]/mol L
–1
0.100 0.0905
0.0820 0.0741 0.0671 0.0549 0.0439 0.0210 0.017
We can determine the difference in concentration over different intervals
of time and thus determine the average rate by dividing ∆[R] by ∆t
(Table 4.1).
{ }
Example 4.1Example 4.1
Example 4.1Example 4.1
Example 4.1
SolutionSolution
SolutionSolution
Solution
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98Chemistry
It can be seen (Table 4.1) that the average rate falls from 1.90× 0
-4
mol L
-1
s
-1
to 0.4 × 10
-4
mol L
-1
s
-1
. However, average rate cannot be used to predict
the rate of a reaction at a particular instant as it would be constant for the
time interval for which it is calculated. So, to express the rate at a particular
moment of time we determine the instantaneous rate. It is obtained when
we consider the average rate at the smallest time interval say dt ( i.e. when
∆t approaches zero). Hence, mathematically for an infinitesimally small
dt instantaneous rate is given by
[
]
[
]
−∆ ∆
= =
∆ ∆
av
R P
r
t t
(4.3)
As ∆t → 0 or
[
]
[
]
inst
d d
R P
d d
r
t t
−
= =
Table 4.1: Average rates of hydrolysis of butyl chloride
[C
4
H
9
CI]
t
1
/ [C
4
H
9
CI]
t
2
/ t
1
/s t
2
/s r
av
× 10
4
/mol L
–1
s
–1
mol L
–1
mol L
–1
[ ] [ ]
( )
{
}
= – − ×
2 1
4
4 9 4 9 2 1
t t
C H Cl – C H Cl / t t 10
0.100 0.0905 0 50 1.90
0.0905
0.0820 50 100 1.70
0.0820
0.0741 100
150 1.58
0.0741
0.0671 150
200 1.40
0.0671
0.0549 200
300 1.22
0.0549
0.0439 300
400 1.10
0.0439 0.0335 400 500 1.04
0.0210 0.017 700 800 0.4
Fig 4.2
Instantaneous rate of
hydrolysis of butyl
chloride(C
4
H
9
Cl)
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Chemical Kinetics
It can be determined graphically by drawing a tangent at time t on
either of the curves for concentration of R and P vs time t and calculating
its slope (Fig. 4.1). So in problem 4.1, r
inst
at 600s for example, can be
calculated by plotting concentration of butyl chloride as a function of
time. A tangent is drawn that touches the curve at t = 600 s (Fig. 4.2).
The slope of this tangent gives the instantaneous rate.
So, r
inst
at 600 s = –
mol L
–1
= 5.12 × 10
–5
mol L
–1
s
–
1
At t = 250 s r
inst
= 1.22 × 10
–4
mol L
–1
s
–1
t = 350 s r
inst
= 1.0 × 10
–4
mol L
–1
s
–1
t = 450 s r
inst
= 6.4 × 10
–5
mol L
–1
s
–1
Now consider a reaction
Hg(l) + Cl
2
(g) →
HgCl
2
(s)
Where stoichiometric coefficients of the reactants and products are
same, then rate of the reaction is given as
[
]
[
]
[
]
2 2
Hg Cl HgCl
Rate of reaction = – –
t t t
∆ ∆ ∆
= =
∆ ∆ ∆
i.e., rate of disappearance of any of the reactants is same as the rate
of appearance of the products. But in the following reaction, two moles of
HI decompose to produce one mole each of H
2
and I
2,
2HI(g) →
H
2
(g) + I
2
(g)
For expressing the rate of such a reaction where stoichiometric
coefficients of reactants or products are not equal to one, rate of
disappearance of any of the reactants or the rate of appearance of
products is divided by their respective stoichiometric coefficients. Since
rate of consumption of HI is twice the rate of formation of H
2
or I
2
, to
make them equal, the term ∆[HI] is divided by 2. The rate of this reaction
is given by
Rate of reaction
[ ]
[
]
[
]
2 2
H I
1
HI
2
t t t
∆ ∆
∆
= − = =
∆ ∆ ∆
Similarly, for the reaction
5 Br
-
(aq) + BrO
3
–
(aq) + 6 H
+
(aq) → 3 Br
2
(aq) + 3 H
2
O (l)
Rate
Br
BrO
H
Br H
= −
[ ]
= −
= −
[ ]
=
[ ]
=
−
−
+
1
5
1
6
1
3
1
3
3 2 2
∆
∆
∆
∆
∆
∆
∆
∆
∆
t t t t
OO
[ ]
∆t
For a gaseous reaction at constant temperature, concentration is
directly proportional to the partial pressure of a species and hence, rate
can also be expressed as rate of change in partial pressure of the reactant
or the product.
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100Chemistry
Intext QuestionsIntext Questions
Intext QuestionsIntext Questions
Intext Questions
4.1 For the reaction R → P, the concentration of a reactant changes from 0.03M
to 0.02M in 25 minutes. Calculate the average rate of reaction using units
of time both in minutes and seconds.
4.2 In a reaction, 2A → Products, the concentration of A decreases from 0.5
mol L
–1
to 0.4 mol L
–1
in 10 minutes. Calculate the rate during this interval?
Rate of reaction depends upon the experimental conditions such
as concentration of reactants (pressure in case of gases),
temperature and catalyst.
The rate of a chemical reaction at a given temperature may depend on
the concentration of one or more reactants and products. The
representation of rate of reaction in terms of concentration of the
reactants is known as rate law. It is also called as rate equation or rate
expression.
The results in Table 4.1 clearly show that rate of a reaction decreases with
the passage of time as the concentration of reactants decrease. Conversely,
rates generally increase when reactant concentrations increase. So, rate of
a reaction depends upon the concentration of reactants.
Example 4.2Example 4.2
Example 4.2Example 4.2
Example 4.2
4.2.2 Rate
Expression
and Rate
Constant
The decomposition of N
2
O
5
in CCl
4
at 318K has been studied by
monitoring the concentration of N
2
O
5
in the solution. Initially the
concentration of N
2
O
5
is 2.33 mol L
–1
and after 184 minutes, it is reduced
to 2.08 mol L
–1
. The reaction takes place according to the equation
2 N
2
O
5
(g) → 4 NO
2
(g) + O
2
(g)
Calculate the average rate of this reaction in terms of hours, minutes
and seconds. What is the rate of production of NO
2
during this period?
Average Rate
=
−
[ ]
= −
−
( )
−
1
2
1
2
2 08 2 33
184
2 5
1
∆
∆
N O
mol L
t
. .
min
= 6.79 × 10
–4
mol L
–1
/min = (6.79 × 10
–4
mol L
–1
min
–1
) × (60 min/1h)
= 4.07 × 10
–2
mol L
–1
/h
= 6.79 × 10
–4
mol L
–1
× 1min/60s
= 1.13 × 10
–5
mol L
–1
s
–1
It may be remembered that
Rate
NO
=
[ ]
1
4
2
∆
∆t
[
]
2
NO
t
∆
=
∆
6.79 × 10
–4
× 4 mol L
–1
min
–1
= 2.72 × 10
–3
mol L
–1
min
–1
SolutionSolution
Solution
Solution
Solution
4.2
4.2
4.2
4.2
4.2
Factors InfluencingFactors Influencing
Factors InfluencingFactors Influencing
Factors Influencing
Rate of a ReactionRate of a Reaction
Rate of a ReactionRate of a Reaction
Rate of a Reaction
4.2.1 Dependence
of Rate on
Concentration
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Chemical Kinetics
Consider a general reaction
aA + bB → cC + dD
where a, b, c and d are the stoichiometric coefficients of reactants
and products.
The rate expression for this reaction is
Rate ∝ [A]
x
[B]
y
(4.4)
where exponents x and y may or may not be equal to the
stoichiometric coefficients (a and b) of the reactants. Above equation
can also be written as
Rate = k [A]
x
[B]
y
(4.4a)
[
]
[ ] [ ]
x y
d
R
A B
d
k
t
− =
(4.4b)
This form of equation (4.4 b) is known as differential rate equation,
where k is a proportionality constant called rate constant
. The equation
like (4.4), which relates the rate of a reaction to concentration of reactants
is called rate law or rate expression. Thus, rate law is the expression
in which reaction rate is given in terms of molar concentration
of reactants with each term raised to some power, which may
or may not be same as the stoichiometric coefficient of the
reacting species in a balanced chemical equation. For example:
2NO(g) + O
2
(g) → 2NO
2
(g)
We can measur
e the rate of this r
eaction as a function of initial
concentrations either by keeping the concentration of one of the reactants
constant and changing the concentration of the other reactant or by
changing the concentration of both the reactants. The following results
are obtained (Table 4.2).
Table 4.2: Initial rate of formation of NO
2
Experiment Initial [NO]/ mol L
-1
Initial [O
2
]/ mol L
-1
Initial rate of
formation of NO
2
/ mol L
-1
s
-1
1. 0.30 0.30 0.096
2. 0.60 0.30 0.384
3. 0.30 0.60 0.192
4. 0.60 0.60 0.768
It is obvious, after looking at the results, that when the concentration
of NO is doubled and that of O
2
is kept constant then the initial rate
increases by a factor of four from 0.096 to 0.384 mol L
–1
s
–1
. This
indicates that the rate depends upon the square of the concentration of
NO. When concentration of NO is kept constant and concentration of
O
2
is doubled the rate also gets doubled indicating that rate depends
on concentration of O
2
to the first power. Hence, the rate equation for
this reaction will be
Rate = k[NO]
2
[O
2
]
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102Chemistry
The differential form of this rate expression is given as
[
]
[ ]
[ ]
2
2
d
R
O
NO
d
k
t
− =
Now, we observe that for this reaction in the rate equation derived
from the experimental data, the exponents of the concentration terms
are the same as their stoichiometric coefficients in the balanced
chemical equation.
Some other examples are given below:
Reaction Experimental rate expression
1. CHCl
3
+ Cl
2
→ CCl
4
+ HCl Rate = k [CHCl
3
] [Cl
2
]
1/2
2. CH
3
COOC
2
H
5
+ H
2
O → CH
3
COOH + C
2
H
5
OH Rate = k [CH
3
COOC
2
H
5
]
1
[H
2
O]
0
In these reactions, the exponents of the concentration terms are not
the same as their stoichiometric coefficients. Thus, we can say that:
Rate law for any reaction cannot be predicted by merely looking
at the balanced chemical equation, i.e., theoretically but must be
determined experimentally.
In the rate equation (4.4)
Rate = k [A]
x
[B]
y
x and y indicate how sensitive the rate is to the change in concentration
of A and B. Sum of these exponents, i.e., x + y in (4.4) gives the overall
order of a reaction whereas x and y represent the order with respect
to the reactants A and B respectively.
Hence, the sum of powers of the concentration of the reactants
in the rate law expression is called the order of that chemical
reaction.
Order of a reaction can be 0, 1, 2, 3 and even a fraction. A zero
order reaction means that the rate of reaction is independent of the
concentration of reactants.
4.2.3 Order of a
Reaction
Calculate the overall order of a reaction which
has the rate expression
(a) Rate = k [A]
1/2
[B]
3/2
(b) Rate =
k [A]
3/2
[B]
–1
(a) Rate = k [A]
x
[B]
y
order = x + y
So order = 1/2 + 3/2 = 2, i.e., second order
(b) order = 3/2 + (–1) = 1/2, i.e., half or
der.
Example 4.3Example 4.3
Example 4.3Example 4.3
Example 4.3
SolutionSolution
SolutionSolution
Solution
A balanced chemical equation never gives us a true picture of how
a reaction takes place since rarely a reaction gets completed in one
step. The reactions taking place in one step are called elementary
reactions. When a sequence of elementary reactions (called mechanism)
gives us the products, the reactions are called complex reactions.
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Chemical Kinetics
Example 4.4Example 4.4
Example 4.4Example 4.4
Example 4.4
SolutionSolution
SolutionSolution
Solution
These may be consecutive reactions (e.g., oxidation of ethane to CO
2
and H
2
O passes through a series of intermediate steps in which alcohol,
aldehyde and acid are formed), reverse reactions and side reactions
(e.g., nitration of phenol yields o-nitrophenol and p-nitrophenol).
Units of rate constant
For a general reaction
aA + bB → cC + dD
Rate = k [A]
x
[B]
y
Where x + y = n = order of the reaction
k =
x
Rate
[A] [B]
y
( )
( )
=
n
concentration 1
= × where [A] [B]
time
concentration
Taking SI units of concentration, mol L
–1
and time, s, the units of
k for different reaction order are listed in Table 4.3
Table 4.3: Units of rate constant
Reaction Order Units of rate constant
Zero order reaction 0
( )
1
1 1
0
1
mol L
1
mol L s
s
mol L
−
− −
−
× =
First order reaction 1
( )
1
1
1
1
mol L
1
s
s
mol L
−
−
−
× =
Second order reaction 2
( )
1
1 1
2
1
mol L
1
mol L s
s
mol L
−
− −
−
× =
Identify the reaction order from each of the following rate constants.
(i) k = 2.3 × 10
–5
L mol
–1
s
–1
(ii) k = 3 × 10
–4
s
–1
(i) The unit of second order rate constant is L mol
–1
s
–1
, therefore
k = 2.3 × 10
–5
L mol
–1
s
–1
represents a second order reaction.
(ii) The unit of a first order rate constant is s
–1
therefore
k = 3 × 10
–4
s
–1
represents a first order reaction.
4.2.4 Molecularity
of a
Reaction
Another property of a reaction called molecularity helps in
understanding its mechanism. The number of reacting species
(atoms, ions or molecules) taking part in an elementary reaction,
which must collide simultaneously in order to bring about a
chemical reaction is called molecularity of a reaction. The reaction
can be unimolecular when one reacting species is involved, for example,
decomposition of ammonium nitrite.
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104Chemistry
NH
4
NO
2
→ N
2
+ 2H
2
O
Bimolecular reactions involve simultaneous collision between two
species, for example, dissociation of hydrogen iodide.
2HI → H
2
+ I
2
Trimolecular or termolecular reactions involve simultaneous collision
between three reacting species, for example,
2NO + O
2
→ 2NO
2
The probability that more than three molecules can collide and
react simultaneously is very small. Hence, reactions with the
molecularity three are very rare and slow to proceed.
It is, therefore, evident that complex reactions involving more than
three molecules in the stoichiometric equation must take place in more
than one step.
KClO
3
+ 6FeSO
4
+ 3H
2
SO
4
→ KCl + 3Fe
2
(SO
4
)
3
+ 3H
2
O
This reaction which apparently seems to be of tenth order is actually
a second order reaction. This shows that this reaction takes place in
several steps. Which step controls the rate of the overall reaction? The
question can be answered if we go through the mechanism of reaction,
for example, chances to win the relay race competition by a team
depend upon the slowest person in the team. Similarly, the overall rate
of the reaction is controlled by the slowest step in a reaction called the
rate determining step. Consider the decomposition of hydrogen
peroxide which is catalysed by iodide ion in an alkaline medium.
2H
2
O
2
-
I
Alkaline medium
→
2H
2
O + O
2
The rate equation for this reaction is found to be
[
]
[ ]
[
]
2 2
2 2
d H O
Rate H O
I
d
−
−
= = k
t
This reaction is first order with respect to both H
2
O
2
and I
–
.
Evidences
suggest that this reaction takes place in two steps
(1) H
2
O
2
+
I
–
→ H
2
O
+
IO
–
(2) H
2
O
2
+
IO
–
→ H
2
O
+
I
–
+
O
2
Both the steps are bimolecular elementary reactions. Species IO
-
is
called as an intermediate since it is formed during the course of the
reaction but not in the overall balanced equation. The first step, being
slow, is the rate determining step. Thus, the rate of formation of
intermediate will determine the rate of this reaction.
Thus, from the discussion, till now, we conclude the following:
(i) Order of a reaction is an experimental quantity. It can be zero and
even a fraction but molecularity cannot be zero or a non integer.
(ii) Order is applicable to elementary as well as complex reactions
whereas molecularity is applicable only for elementary reactions.
For complex reaction molecularity has no meaning.
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Chemical Kinetics
(iii) For complex reaction, order is given by the slowest step and
molecularity of the slowest step is same as the order of the overall
reaction.
Intext QuestionsIntext Questions
Intext QuestionsIntext Questions
Intext Questions
4.3 For a reaction, A + B → Product; the rate law is given by,
r = k [ A]
1/2
[B]
2
.
What is the order of the reaction?
4.4 The conversion of molecules X to Y follows second order kinetics. If
concentration of X is increased to three times how will it affect the rate of
formation of Y ?
We have already noted that the concentration dependence of rate is
called differential rate equation. It is not always convenient to
determine the instantaneous rate, as it is measured by determination
of slope of the tangent at point ‘t’ in concentration vs time plot
(Fig.
4.1). This makes it difficult to determine the rate law and hence
the order of the reaction. In order to avoid this difficulty, we can
integrate the differential rate equation to give a relation between directly
measured experimental data, i.e., concentrations at different times
and rate constant.
The integrated rate equations are different for the reactions of different
reaction orders. We shall determine these equations only for zero and
first order chemical reactions.
Zero order reaction means that the rate of the reaction is proportional
to zero power of the concentration of reactants. Consider the reaction,
R → P
Rate =
[
]
[ ]
0
d
R
R
d
k
t
− =
As any quantity raised to power zero is unity
Rate =
[
]
d
R
1
d
k ×
t
− =
d[R] = – k dt
Integrating both sides
[R] = – k t + I (4.5)
where, I is the constant of integration.
At t = 0, the concentration of the reactant R = [R]
0
, where [R]
0
is
initial concentration of the reactant.
Substituting in equation (4.5)
[R]
0
= –k × 0 + I
[R]
0
= I
Substituting the value of I in the equation (4.5)
[R] = -kt + [R]
0
(4.6)
4.34.3
4.34.3
4.3
IntegratedIntegrated
IntegratedIntegrated
Integrated
RateRate
RateRate
Rate
EquationsEquations
EquationsEquations
Equations
4.3.1 Zero Order
Reactions
2020-21
106Chemistry
Fig. 4.3: Variation in the concentration vs
time plot for a zero order reaction
Time
k = -slope
Concentration of R
[R ]
0
0
Comparing (4.6) with equation of a straight line,
y = mx + c, if we plot [R] against t, we get a straight
line (Fig. 4.3) with slope = –k and intercept equal
to [R]
0
.
Further simplifying equation (4.6), we get the rate
constant, k as
[
]
[
]
0
R R
k
t
−
=
(4.7)
Zero order reactions are relatively uncommon but
they occur under special conditions. Some enzyme
catalysed reactions and reactions which occur on
metal surfaces are a few examples of zero order
reactions. The decomposition of gaseous ammonia
on a hot platinum surface is a zero order reaction at
high pressure.
(
)
(
)
(
)
1130K
3 2 2
Pt catalyst
2NH g N g +3H g
→
Rate = k [NH
3
]
0
= k
In this reaction, platinum metal acts as a catalyst. At high pressure,
the metal surface gets saturated with gas molecules. So, a further
change in reaction conditions is unable to alter the amount of ammonia
on the surface of the catalyst making rate of the reaction independent
of its concentration. The thermal decomposition of HI on gold surface
is another example of zero order reaction.
In this class of reactions, the rate of the reaction is proportional to the
first power of the concentration of the reactant R. For example,
R → P
[
]
[ ]
d
R
Rate
R
d
k
t
= − =
or
[
]
[ ]
d R
– d
R
k t
=
Integrating this equation, we get
ln [R] = –kt + I (4.8)
Again, I is the constant of integration and its value can be determined
easily.
When t = 0, R = [R]
0
, where [R]
0
is the initial concentration of the
reactant.
Therefore, equation (4.8) can be written as
ln [R]
0
= –k × 0 + I
ln [R]
0
= I
Substituting the value of I in equation (4.8)
ln[R] = -kt + ln[R]
0
(4.9)
4.3.2 First Order
Reactions
2020-21
107
Chemical Kinetics
Rearranging this equation
[
]
[ ]
0
R
ln
R
kt
= −
or
[
]
[ ]
0
R
1
ln
R
=k
t
(4.10)
At time t
1
from equation (4.8)
*ln[R]
1
= –
kt
1
+ *ln[R]
0
(4.11)
At time t
2
ln[R]
2
= – kt
2
+ ln[R]
0
(4.12)
where, [R]
1
and [R]
2
are the concentrations of the reactants at time
t
1
and
t
2
respectively.
Subtracting (4.12) from (4.11)
ln[R]
1
– ln[R]
2
= –
kt
1
– (–kt
2
)
[
]
[ ]
( )
1
2 1
2
R
ln k
R
t t
−
=
( )
[
]
[ ]
1
2 1
2
1
R
ln
R
k
t t
=
−
(4.13)
Equation (4.9) can also be written as
[
]
[ ]
= −
0
R
ln
R
kt
Taking antilog of both sides
[R] = [R]
0
e
-kt
(4.14)
Comparing equation (4.9) with y = mx + c, if we plot ln [R] against
t (Fig. 4.4) we get a straight line with slope = –k and intercept equal to
ln [R]
0
The first order rate equation (4.10) can also be written in the form
[
]
[ ]
0
R
2.303
log
R
k
t
=
(4.15)
*
[
]
[ ]
0
R
log
2.303
R
kt
=
If we plot a graph between log [R]
0
/[R] vs t, (Fig. 4.5),
the slope = k/2.303
Hydrogenation of ethene is an example of first order reaction.
C
2
H
4
(g) + H
2
(g) → C
2
H
6
(g)
Rate = k [C
2
H
4
]
All natural and artificial radioactive decay of unstable nuclei take
place by first order kinetics.
* Refer to Appendix-IV for ln and log (logarithms).
2020-21
108Chemistry
Fig. 4.4: A plot between ln[R] and t
for a first order reaction
Fig. 4.5: Plot of log [R]
0
/[R] vs time for a first
order reaction
Slope = /2.303k
log ([R /[R])
0
]
Time
0
226 4 222
88 2 86
Ra He Rn
→ +
Rate = k [Ra]
Decomposition of N
2
O
5
and N
2
O are some more examples of first
order reactions.
The initial concentration of N
2
O
5
in the following first order reaction
N
2
O
5
(g) → 2 NO
2
(g) + 1/2O
2
(g) was 1.24 × 10
–2
mol L
–1
at 318 K. The
concentration of N
2
O
5
after 60 minutes was 0.20 × 10
–2
mol L
–1
.
Calculate the rate constant of the reaction at 318 K.
For a first order reaction
[
]
[ ]
1
2
R
log
R
=
(
)
2 1
2.303
k t t
−
k =
( )
[
]
[ ]
1
2 1
2
2.303
log
R
t t
R
−
=
( )
2 1
2 1
1.24 10 mol L
2.303
log
60 min 0 min
0.20 10 mol L
− −
− −
×
−
×
=
1
2.303
log 6.2 min
60
−
k = 0.0304 min
-1
Example 4.5Example 4.5
Example 4.5Example 4.5
Example 4.5
SolutionSolution
SolutionSolution
Solution
Let us consider a typical first order gas phase reaction
A(g) → B(g) + C(g)
Let p
i
be the initial pressure of A and p
t
the total pressure at
time ‘t’. Integrated rate equation for such a reaction can be derived as
Total pressure p
t
= p
A
+ p
B
+ p
C
(pressure units)
2020-21
109
Chemical Kinetics
p
A
, p
B
and p
C
are the partial pressures of A, B and C, respectively.
If x atm be the decrease in pressure of A at time t and one mole each
of B and C is being formed, the increase in pressure of B and C will also
be x atm each.
A(g) → B(g) + C(g)
At t = 0 p
i
atm 0 atm 0 atm
At time t (p
i
–x) atm x atm x atm
where, p
i
is the initial pressure at time t = 0.
p
t
= (p
i
– x) + x + x = p
i
+ x
x = (p
t
- p
i
)
where, p
A
= p
i
– x = p
i
– (p
t
– p
i
)
= 2p
i
– p
t
k =
i
A
2.303
log
p
p
t
(4.16)
=
( )
i
i t
2.303
log
2
p
p p
t −
The following data were obtained during the first order thermal
decomposition of N
2
O
5
(g) at constant volume:
(
)
(
)
(
)
2 5 2 4 2
g g g
2N O 2N O O→ +
S.No. Time/s Total Pressure/(atm)
1. 0 0.5
2. 100 0.512
Calculate the rate constant.
Let the pressure of N
2
O
5
(g) decrease by 2x atm. As two moles of
N
2
O
5
decompose to give two moles of N
2
O
4
(g) and one mole of O
2
(g),
the pressure of N
2
O
4
(g) increases by 2x atm and that of O
2
(g)
increases by x atm.
(
)
(
)
(
)
2 5 2 4 2
g g g
2N O 2N O O→ +
Start t = 0 0.5 atm 0 atm 0 atm
At time t (0.5 – 2x) atm 2x atm x atm
p
t
=
2 5 2 4 2
N O N O O
p p p
+ +
= (0.5 – 2x) + 2x + x = 0.5 + x
x 0.5
t
p
= −
2 5
N O
p
= 0.5 – 2x
= 0.5 – 2 (p
t
– 0.5) = 1.5 – 2p
t
At t = 100 s; p
t
= 0.512 atm
Example 4.6Example 4.6
Example 4.6Example 4.6
Example 4.6
SolutionSolution
SolutionSolution
Solution
2020-21
110Chemistry
The half-life of a reaction is the time in which the concentration of a
reactant is reduced to one half of its initial concentration. It is
represented as t
1/2
.
For a zero order reaction, rate constant is given by equation 4.7.
[
]
[
]
0
R R
k
t
−
=
[ ]
[ ]
0
1/2
1
R
At ,
R
2
t t= =
The rate constant at t
1/2
becomes
[
]
[
]
0 0
1/2
1/2
R R
k
t
−
=
[
]
0
1/ 2
R
2
t
k
=
It is clear that t
1/2
for a zero order reaction is directly proportional
to the initial concentration of the reactants and inversely proportional
to the rate constant.
For the first order reaction,
[
]
[ ]
0
2.303
R
log
R
k
t
=
(4.15)
at t
1/2
[ ]
[
]
0
R
R
2
=
(4.16)
So, the above equation becomes
[
]
[ ]
0
1/ 2
2.303
R
log
/2
R
k
t
=
or
1/2
2.303
log 2
t
k
=
1/2
2.303
0.301
t
k
= ×
1/ 2
0.693
t
k
=
(4.17)
2 5
N O
p
= 1.5 – 2 × 0.512 = 0.476 atm
Using equation (4.16)
i
A
4 1
0.5 atm
2.303 2.303
log log
100 s 0.476 atm
2.303
0.0216 4.98 10 s
100 s
p
k
t p
− −
= =
= × = ×
4.3.3 Half-Life of a
Reaction
2020-21
111
Chemical Kinetics
A first order reaction is found to have a rate constant, k
= 5.5 × 10
-14
s
-1
.
Find the half-life of the reaction.
Half-life for a first order reaction is
1/ 2
t
=
0.693
k
1/ 2
t
=
–14 –1
0.693
5.5 ×10 s
= 1.26 × 10
13
s
Show that in a first order reaction, time required for completion of
99.9% is 10 times of half-life (t
1/2
) of the reaction.
When reaction is completed 99.9%, [R]
n
= [R]
0
– 0.999[R]
0
k =
[
]
[ ]
0
2.303
R
log
R
t
=
[
]
[ ] [ ]
0
0 0
2.303
R
log
0.999
R R
t −
=
3
2.303
log10
t
t = 6.909/k
For half-life of the reaction
t
1/2
= 0.693/k
1/2
t
t
=
6.909
10
0.693
k
k
× =
It can be seen that for a first order reaction, half-life period is
constant, i.e., it is independent of initial concentration of the reacting
species. The half-life of a first order equation is readily calculated from
the rate constant and vice versa.
For zero order reaction t
1/2
∝∝
∝∝
∝ [R]
0
. For first order reaction
t
1/2
is independent of [R]
0
.
Example 4.7Example 4.7
Example 4.7
Example 4.7
Example 4.7
SolutionSolution
Solution
Solution
Solution
Example 4.8Example 4.8
Example 4.8Example 4.8
Example 4.8
SolutionSolution
SolutionSolution
Solution
Table 4.4 summarises the mathematical features of integrated laws of
zero and first order reactions.
Table 4.4: Integrated Rate Laws for the Reactions of Zero and First Order
Order Reaction Differential Integrated Straight Half- Units of k
type rate law rate law line plot life
0 R→ P d[R]/dt = -k kt = [R]
0
-[R] [R] vs t [R]
0
/2k conc time
-1
or mol L
–1
s
–1
1 R→ P d[R]/dt = -k[R] [R] = [R]
0
e
-kt
ln[R] vs t ln 2/k time
-1
or s
–1
or kt =
ln{[R]
0
/[R]}
2020-21
112Chemistry
Most of the chemical reactions are accelerated by increase in temperature.
For example, in decomposition of N
2
O
5
, the time taken for half of the
original amount of material to decompose is 12 min at 50
o
C, 5 h at
25
o
C and 10 days at 0
o
C. You also know that in a mixture of potassium
permanganate (KMnO
4
) and oxalic acid (H
2
C
2
O
4
), potassium
permanganate gets decolourised faster at a higher temperature than
that at a lower temperature.
It has been found that for a chemical reaction with rise in
temperature by 10°, the rate constant is nearly doubled.
The temperature dependence of the rate of a chemical reaction can
be accurately explained by Arrhenius equation (4.18). It was first
proposed by Dutch chemist, J.H. van’t Hoff but Swedish chemist,
Arrhenius provided its physical justification and interpretation.
4.4 Temperature4.4 Temperature
4.4 Temperature4.4 Temperature
4.4 Temperature
Dependence ofDependence of
Dependence ofDependence of
Dependence of
the Rate of athe Rate of a
the Rate of athe Rate of a
the Rate of a
ReactionReaction
ReactionReaction
Reaction
The order of a reaction is sometimes altered by conditions. There
are many reactions which obey first order rate law although they are
higher order reactions. Consider the hydrolysis of ethyl acetate which
is a chemical reaction between ethyl acetate and water. In reality, it
is a second order reaction and concentration of both ethyl acetate and
water affect the rate of the reaction. But water is taken in large excess
for hydrolysis, therefore, concentration of water is not altered much
during the reaction. Thus, the rate of reaction is affected by
concentration of ethyl acetate only. For example, during the hydrolysis
of 0.01 mol of ethyl acetate with 10 mol of water, amounts of the
reactants and products at the beginning (t = 0) and completion (t) of
the reaction are given as under.
CH
3
COOC
2
H
5
+ H
2
O
H
+
→
CH
3
COOH + C
2
H
5
OH
t = 0 0.01 mol 10 mol 0 mol
0 mol
t 0 mol 9.99 mol 0.01 mol
0.01 mol
The concentration of water does not get altered much during the
course of the reaction. So,
the reaction behaves as first order reaction.
Such reactions are called pseudo first order reactions.
Inversion of cane sugar is another pseudo first order reaction.
C
12
H
22
O
11
+ H
2
O
+
H
→
C
6
H
12
O
6
+ C
6
H
12
O
6
Cane sugar Glucose Fructose
Rate = k [C
12
H
22
O
11
]
Intext QuestionsIntext Questions
Intext QuestionsIntext Questions
Intext Questions
4.5 A first order reaction has a rate constant 1.15 × 10
-3
s
-1
. How long will 5 g of
this reactant take to reduce to 3 g?
4.6 Time required to decompose SO
2
Cl
2
to half of its initial amount is 60
minutes. If the decomposition is a first order reaction, calculate the rate
constant of the reaction.
2020-21
113
Chemical Kinetics
k = A e
-Ea /RT
(4.18)
where A is the Arrhenius factor or the frequency factor. It is also called
pre-exponential factor. It is a constant specific to a particular reaction.
R is gas constant and E
a
is activation energy measured in joules/mole
(J mol
–1
).
It can be understood clearly using the following simple reaction
(
)
(
)
(
)
2 2
H g I g 2HI g+ →
According to Arrhenius, this reaction can take place
only when a molecule of hydrogen and a molecule of iodine
collide to form an unstable intermediate (Fig. 4.6). It exists
for a very short time and then breaks up to form two
molecules of hydrogen iodide.
Fig. 4.6: Formation of HI through
the intermediate
Intermediate
Fig. 4.7: Diagram showing plot of potential
energy vs reaction coordinate
Fig. 4.8: Distribution curve showing energies
among gaseous molecules
The energy required to form this
intermediate, called activated complex
(C), is known as activation energy (E
a
).
Fig. 4.7 is obtained by plotting potential
energy vs reaction coordinate. Reaction
coordinate represents the profile of energy
change when reactants change into
products.
Some energy is released when the
complex decomposes to form products.
So, the final enthalpy of the reaction
depends upon the nature of reactants
and products.
All the molecules in the reacting
species do not have the same kinetic
energy. Since it is difficult to predict the
behaviour of any one molecule with
precision, Ludwig Boltzmann and James
Clark Maxwell used statistics to predict
the behaviour of large number of
molecules. According to them, the
distribution of kinetic energy may be
described by plotting the fraction of
molecules (N
E
/N
T
) with a given kinetic
energy (E) vs kinetic energy (Fig. 4.8).
Here, N
E
is the number of molecules with
energy E and N
T
is total number
of molecules.
The peak of the curve corresponds to
the most probable kinetic energy, i.e.,
kinetic energy of maximum fraction of
molecules. There are decreasing number
of molecules with energies higher or
lower than this value. When the
2020-21
114Chemistry
Fig. 4.10: A plot between ln k and 1/T
In Fig. 4.10, slope = –
a
E
R
and intercept = ln
A.
So we can calculate E
a
and A using these values.
At temperature T
1
, equation (4.19) is
ln k
1
= –
a
1
E
RT
+ ln A (4.20)
At temperature T
2
, equation (4.19) is
ln k
2
= –
a
2
E
RT
+ ln A (4.21)
(since A is constant for a given reaction)
k
1
and k
2
are the values of rate constants at
temperatures T
1
and T
2
respectively.
Fig. 4.9: Distribution curve showing temperature
dependence of rate of a reaction
temperature is raised, the maximum
of the curve moves to the higher
energy value (Fig. 4.9) and the curve
broadens out, i.e., spreads to the right
such that there is a greater proportion
of molecules with much higher
energies. The area under the curve
must be constant since total
probability must be one at all times.
We can mark the position of E
a
on
Maxwell Boltzmann distribution curve
(Fig. 4.9).
Increasing the temperature of the substance increases the fraction
of molecules, which collide with energies greater than E
a
. It is clear
from the diagram that in the curve at (t + 10), the area showing the
fraction of molecules having energy equal to or greater than activation
energy gets doubled leading to doubling the rate of a reaction.
In the Arrhenius equation (4.18) the factor e
-Ea /RT
corresponds to
the fraction of molecules that have kinetic energy greater than E
a
.
Taking natural logarithm of both sides of equation (4.18)
ln k = –
a
E
RT
+ ln A (4.19)
The plot of ln k vs 1/T gives a straight line according to the equation
(4.19) as shown in Fig. 4.10.
Thus, it has been found from Arrhenius equation (4.18) that
increasing the temperature or decreasing the activation energy will
result in an increase in the rate of the reaction and an exponential
increase in the rate constant.
2020-21
115
Chemical Kinetics
The rate constants of a reaction at 500K and 700K are 0.02s
–1
and
0.07s
–1
respectively. Calculate the values of E
a
and A.
2
1
log
k
k
=
2 1
a
1 2
2.303
T T
E
T T
R
−
0.07
log
0.02
=
a
1 1
700 500
2.303 8.314 J mol
700 500
E
K
− −
−
×
×
0.544 = E
a
× 5.714 × 10
-4
/19.15
E
a
= 0.544 × 19.15/5.714 × 10
–4
= 18230.8 J
Since k = Ae
-Ea/RT
0.02 = Ae
-18230.8/8.314 × 500
A = 0.02/0.012 = 1.61
The first order rate constant for the decomposition of ethyl iodide
by the reaction
C
2
H
5
I(g) → C
2
H
4
(g) + HI(g)
at 600K is 1.60 × 10
–5
s
–1
. Its energy of activation is 209 kJ/mol.
Calculate the rate constant of the reaction at 700K.
We know that
log k
2
– log k
1
=
a
1 2
1 1
2.303
E
T T
R
−
Subtracting equation (4.20) from (4.21), we obtain
ln k
2
–
ln k
1
=
a
1
E
RT
–
a
2
E
RT
ln
k
k
E
R
T T
2
1
1 2
1 1
=
−
a
log
.
k
k
E
R
T T
2
1
1 2
2 303
1 1
=
−
a
(4.22)
log
.
k
k
E
T T
T T
2
1
2 1
1 2
2 303
=
−
a
R
SolutionSolution
SolutionSolution
Solution
SolutionSolution
SolutionSolution
Solution
Example 4.10Example 4.10
Example 4.10Example 4.10
Example 4.10
Example 4.9Example 4.9
Example 4.9Example 4.9
Example 4.9
2020-21
116Chemistry
A catalyst is a substance which increases the rate of a reaction without
itself undergoing any permanent chemical change. For example, MnO
2
catalyses the following reaction so as to increase its rate considerably.
2KClO
3
MnO
2
→
2 KCl + 3O
2
The word catalyst should not be used when the added substance
reduces the rate of raction. The substance is then called inhibitor. The
action of the catalyst can be explained by intermediate complex theory.
According to this theory, a catalyst participates in a chemical reaction by
forming temporary bonds with the reactants resulting in an intermediate
complex. This has a transitory existence and decomposes to yield products
and the catalyst.
It is believed that the catalyst provides an
alternate pathway or reaction mechanism by
reducing the activation energy between
reactants and products and hence lowering
the potential energy barrier as shown in
Fig. 4.11.
It is clear from Arrhenius equation (4.18)
that lower the value of activation energy faster
will be the rate of a reaction.
A small amount of the catalyst can catalyse
a large amount of reactants. A catalyst does
not alter Gibbs energy, ∆G of a reaction. It
catalyses the spontaneous reactions but does
not catalyse non-spontaneous reactions. It is
also found that a catalyst does not change the equilibrium constant of
a reaction rather, it helps in attaining the equilibrium faster, that is, it
catalyses the forward as well as the backward reactions to the same
extent so that the equilibrium state remains same but is reached earlier.
Though Arrhenius equation is applicable under a wide range of
circumstances, collision theory, which was developed by Max Trautz
and William Lewis in 1916 -18, provides a greater insight into the
energetic and mechanistic aspects of reactions. It is based on kinetic
theory of gases. According to this theory, the reactant molecules are
4.4.1 Effect of
Catalyst
4.5 Collision4.5 Collision
4.5 Collision4.5 Collision
4.5 Collision
Theory ofTheory of
Theory ofTheory of
Theory of
ChemicalChemical
ChemicalChemical
Chemical
Reactions
Reactions
ReactionsReactions
Reactions
Fig. 4.11:Fig. 4.11:
Fig. 4.11:Fig. 4.11:
Fig. 4.11:
Effect of catalyst on activation
energy
log k
2
=
a
1
1 2
1 1
log
2.303
E
k
T T
R
−
+
=
( )
1
5
1 1
1 1
209000 J mol L
log
1.60 10
600 K 700 K
2.303 8.314 J mol L K
−
−
− −
−
+
×
×
log k
2
= – 4.796 + 2.599 = – 2.197
k
2
=
6.36 × 10
–3
s
–1
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Chemical Kinetics
assumed to be hard spheres and reaction is postulated to occur when
molecules collide with each other. The number of collisions per
second per unit volume of the reaction mixture is known as
collision frequency (Z). Another factor which affects the rate of
chemical reactions is activation energy (as we have already studied).
For a bimolecular elementary reaction
A + B → Products
rate of reaction can be expressed as
a
/
AB
Rate Z e
E RT
−
=
(4.23)
where Z
AB
represents the collision frequency of reactants, A and B
and e
-Ea /R
T
represents the fraction of molecules with energies equal to
or greater than E
a
. Comparing (4.23) with Arrhenius equation, we can
say that A is related to collision frequency.
Equation (4.23) predicts the value of rate constants fairly
accurately for the reactions that involve atomic species or simple
molecules but for complex molecules significant deviations are
observed. The reason could be that all collisions do not lead to the
formation of products. The collisions in which molecules collide with
sufficient kinetic energy (called threshold energy*) and proper
orientation, so as to facilitate breaking of bonds between reacting
species and formation of new bonds to form products are called as
effective collisions.
For example, formation of
methanol from bromoethane depends
upon the orientation of reactant
molecules as shown in
Fig. 4.12. The proper orientation of
reactant molecules lead to bond
formation whereas improper
orientation makes them simply
bounce back and no products are
formed.
To account for effective collisions,
another factor P, called the probability
or steric factor is introduced. It takes into account the fact that in a
collision, molecules must be properly oriented i.e.,
a
/
AB
Rate Z e
E RT
P
−
=
Thus, in collision theory activation energy and proper orientation of
the molecules together determine the criteria for an effective collision
and hence the rate of a chemical reaction.
Collision theory also has certain drawbacks as it considers atoms/
molecules to be hard spheres and ignores their structural aspect. You
will study details about this theory and more on other theories in your
higher classes.
* Threshold energy = Activation Energy + energy possessed by reacting species.
Fig. 4.12:Fig. 4.12:
Fig. 4.12:Fig. 4.12:
Fig. 4.12:
Diagram showing molecules having proper and
improper orientation
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118Chemistry
Intext QuestionsIntext Questions
Intext QuestionsIntext Questions
Intext Questions
4.7 What will be the effect of temperature on rate constant ?
4.8 The rate of the chemical reaction doubles for an increase of 10K in absolute
temperature from 298K. Calculate E
a
.
4.9 The activation energy for the reaction
2 HI(g) → H
2
+ I
2
(g)
is 209.5 kJ mol
–1
at 581K.Calculate the fraction of molecules of reactants
having energy equal to or greater than activation energy?
SummarySummary
SummarySummary
Summary
Chemical kinetics is the study of chemical reactions with respect to reaction
rates, effect of various variables, rearrangement of atoms and formation of
intermediates. The rate of a reaction is concerned with decrease in concentration
of reactants or increase in the concentration of products per unit time. It can
be expressed as instantaneous rate at a particular instant of time and average
rate over a large interval of time. A number of factors such as temperature,
concentration of reactants, catalyst, affect the rate of a reaction. Mathematical
representation of rate of a reaction is given by rate law. It has to be determined
experimentally and cannot be predicted. Order of a reaction with respect to
a reactant is the power of its concentration which appears in the rate law
equation. The order of a reaction is the sum of all such powers of concentration
of terms for different reactants. Rate constant is the proportionality factor in
the rate law. Rate constant and order of a reaction can be determined from rate
law or its integrated rate equation. Molecularity is defined only for an elementary
reaction. Its values are limited from 1 to 3 whereas order can be 0, 1, 2, 3 or
even a fraction. Molecularity and order of an elementary reaction are same.
Temperature dependence of rate constants is described by Arrhenius equation
(k = Ae
–Ea/RT
). E
a
corresponds to the activation energy and is given by the
energy difference between activated complex and the reactant molecules, and A
(Arrhenius factor or pre-exponential factor) corresponds to the collision frequency.
The equation clearly shows that increase of temperature or lowering of E
a
will
lead to an increase in the rate of reaction and presence of a catalyst lowers the
activation energy by providing an alternate path for the reaction. According to
collision theory, another factor P called steric factor which refers to the orientation
of molecules which collide, is important and contributes to effective collisions,
thus, modifying the Arrhenius equation to
a
/
AB
Z e
E RT
k P
−
=
.
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Chemical Kinetics
4.1 From the rate expression for the following reactions, determine their
order of reaction and the dimensions of the rate constants.
(i) 3NO(g) → N
2
O (g) Rate = k[NO]
2
(ii) H
2
O
2
(aq) + 3I
–
(aq) + 2H
+
→ 2H
2
O (l) +
3
I
−
Rate = k[H
2
O
2
][I
-
]
(iii) CH
3
CHO (g) → CH
4
(g) + CO(g) Rate = k[CH
3
CHO]
3/2
(iv) C
2
H
5
Cl (g) → C
2
H
4
(g) + HCl (g) Rate = k[C
2
H
5
Cl]
4.2 For the reaction:
2A + B → A
2
B
the rate = k[A][B]
2
with k = 2.0 × 10
–6
mol
–2
L
2
s
–1
. Calculate the initial
rate of the reaction when [A] = 0.1 mol L
–1
, [B] = 0.2 mol L
–1
. Calculate
the rate of reaction after [A] is reduced to 0.06 mol L
–1
.
4.3 The decomposition of NH
3
on platinum surface is zero order reaction. What
are the rates of production of N
2
and H
2
if k = 2.5 × 10
–4
mol
–1
L
s
–1
?
4.4 The decomposition of dimethyl ether leads to the formation of CH
4
, H
2
and CO and the reaction rate is given by
Rate = k [CH
3
OCH
3
]
3/2
The rate of reaction is followed by increase in pressure in a closed
vessel, so the rate can also be expressed in terms of the partial pressure
of dimethyl ether, i.e.,
(
)
3 3
3 / 2
CH OCH
Rate
p
k=
If the pressure is measured in bar and time in minutes, then what are
the units of rate and rate constants?
4.5 Mention the factors that affect the rate of a chemical reaction.
4.6 A reaction is second order with respect to a reactant. How is the rate
of reaction affected if the concentration of the reactant is
(i) doubled (ii) reduced to half ?
4.7 What is the effect of temperature on the rate constant of a reaction?
How can this effect of temperature on rate constant be represented
quantitatively?
4.8 In a pseudo first order reaction in water, the following results were
obtained:
t/s 0 30 60 90
[A]/ mol L
–1
0.55 0.31 0.17 0.085
Calculate the average rate of reaction between the time interval 30
to 60 seconds.
4.9 A reaction is first order in A and second order in B.
(i) Write the differential rate equation.
(ii) How is the rate affected on increasing the concentration of B three
times?
(iii) How is the rate affected when the concentrations of both A and B
are doubled?
ExercisesExercises
ExercisesExercises
Exercises
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120Chemistry
4.10 In a reaction between A and B, the initial rate of reaction (r
0
) was measured
for different initial concentrations of A and B as given below:
A/ mol L
–1
0.20 0.20 0.40
B/ mol L
–1
0.30 0.10 0.05
r
0
/mol L
–1
s
–1
5.07 × 10
–5
5.07 × 10
–5
1.43 × 10
–4
What is the order of the reaction with respect to A and B?
4.11 The following results have been obtained during the kinetic studies of the reaction:
2A + B → C + D
Experiment [A]/mol L
–1
[B]/mol L
–1
Initial rate of formation
of D/mol L
–1
min
–1
I 0.1 0.1 6.0 × 10
–3
II 0.3 0.2 7.2 × 10
–2
III 0.3 0.4 2.88 × 10
–1
IV 0.4 0.1 2.40 × 10
–2
Determine the rate law and the rate constant for the reaction.
4.12 The reaction between A and B is first order with respect to A and zero order
with respect to B. Fill in the blanks in the following table:
Experiment [A]/ mol L
–1
[B]/ mol L
–1
Initial rate/
mol L
–1
min
–1
I 0.1 0.1 2.0 × 10
–2
II – 0.2 4.0 × 10
–2
III 0.4 0.4 –
IV – 0.2 2.0 × 10
–2
4.13 Calculate the half-life of a first order reaction from their rate constants given
below:
(i) 200 s
–1
(ii) 2 min
–1
(iii) 4 years
–1
4.14 The half-life for radioactive decay of
14
C is 5730 years. An archaeological
artifact containing wood had only 80% of the
14
C found in a living tree. Estimate
the age of the sample.
4.15 The experimental data for decomposition of N
2
O
5
[2N
2
O
5
→ 4NO
2
+ O
2
]
in gas phase at 318K are given below:
t/s 0 400 800 1200
1600 2000 2400 2800 3200
10
2
× [N
2
O
5
]/ 1.63
1.36 1.14 0.93 0.78 0.64 0.53 0.43 0.35
mol L
–1
(i) Plot [N
2
O
5
] against t.
(ii) Find the half-life period for the reaction.
(iii) Draw a graph between log[N
2
O
5
] and t.
(iv) What is the rate law ?
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121
Chemical Kinetics
(v) Calculate the rate constant.
(vi) Calculate the half-life period from k and compare it with (ii).
4.16 The rate constant for a first order reaction is 60 s
–1
. How much time will
it take to reduce the initial concentration of the reactant to its 1/16
th
value?
4.17 During nuclear explosion, one of the products is
90
Sr with half-life of
28.1 years. If 1µg of
90
Sr was absorbed in the bones of a newly born
baby instead of calcium, how much of it will remain after 10 years and
60 years if it is not lost metabolically.
4.18 For a first order reaction, show that time required for 99% completion
is twice the time required for the completion of 90% of reaction.
4.19 A first order reaction takes 40 min for 30% decomposition. Calculate t
1/2
.
4.20 For the decomposition of azoisopropane to hexane and nitrogen at 543
K, the following data are obtained.
t (sec) P(mm of Hg)
0 35.0
360 54.0
720 63.0
Calculate the rate constant.
4.21 The following data were obtained during the first order thermal
decomposition of SO
2
Cl
2
at a constant volume.
(
)
(
)
(
)
2 2 2 2
SO Cl g SO g Cl g→ +
Experiment Time/s
–1
Total pressure/atm
1 0 0.5
2 100 0.6
Calculate the rate of the reaction when total pressure is 0.65 atm.
4.22 The rate constant for the decomposition of N
2
O
5
at various temperatures
is given below:
T/°C 0 20 40 60
80
10
5
× k/s
-1
0.0787 1.70 25.7 178 2140
Draw a graph between ln k and 1/T and calculate the values of A and
E
a
. Predict the rate constant at 30° and 50°C.
4.23 The rate constant for the decomposition of hydrocarbons is 2.418 × 10
–5
s
–
1
at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the
value of pre-exponential factor.
4.24 Consider a certain reaction A → Products with k = 2.0 × 10
–2
s
–1
. Calculate
the concentration of A remaining after 100 s if the initial concentration
of A is 1.0 mol L
–1
.
4.25 Sucrose decomposes in acid solution into glucose and fructose according
to the first order rate law, with t
1/2
= 3.00 hours. What fraction of sample
of sucrose remains after 8 hours ?
4.26 The decomposition of hydrocarbon follows the equation
k = (4.5 × 10
11
s
–1
) e
-28000K/T
Calculate E
a
.
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122Chemistry
4.27 The rate constant for the first order decomposition of H
2
O
2
is given by the
following equation:
log k = 14.34 – 1.25 × 10
4
K/T
Calculate E
a
for this reaction and at what temperature will its half-period
be 256 minutes?
4.28 The decomposition of A into product has value of k as 4.5 × 10
3
s
–1
at 10°C
and energy of activation 60 kJ mol
–1
. At what temperature would k be
1.5 × 10
4
s
–1
?
4.29 The time required for 10% completion of a first order reaction at 298K is
equal to that required for its 25% completion at 308K. If the value of A is
4 × 10
10
s
–1
. Calculate k at 318K and E
a
.
4.30 The rate of a reaction quadruples when the temperature changes from
293 K to 313 K. Calculate the energy of activation of the reaction assuming
that it does not change with temperature.
Answers to Some Intext Questions
4.1 r
av
= 6.66 × 10
–6
Ms
–1
4.2 Rate of reaction = rate of diappearance of A
= 0.005 mol litre
–1
min
–1
4.3 Order of the reaction is 2.5
4.4 X → Y
Rate = k[X]
2
The rate will increase 9 times
4.5 t = 444 s
4.6 1.925 × 10
–4
s
–1
4.8 Ea = 52.897 kJ mol
–1
4.9 1.471 × 10
–19
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