Chemistry
Chemical Kinetics – 1
Session Objectives
•
Rate of reaction
•
Factors affecting rate constant
•
Molecularity
•
Order of a reaction
•
Kinetics of zero order reaction
•
Kinetics of first order reaction
•
Pseudo first order reaction
Rate of reaction
For a reaction,
A  B D
The rate may be expressed as
d[A]
d[B]
d[D]
or 
or
dt
dt
dt
Change in the concentration of species
Rate of reaction =
Time int erval
1
1
Unit of rate of reaction = molL s
Rate of reaction
For general reaction
aA  bB 
 cC  dD
Rate of reaction

1 d(A) 1 d(B) 1 d(C) 1 d(D)



a dt
b dt
c dt
d dt
According to law of mass action
Rate  k  A  B
a
b
Where k is rate constant.
Rate constant
Measure of rate of reaction
Different for different reactions
Independent of concentration of reactants
Depends on the temperature
Graphical representation of rate
• The slope of graph at any instant
gives us the rate of reaction at
that moment and is called
instantaneous rate of the reaction.
• The average of these slopes
over a period gives us rate of
the reaction which is called as
average rate of the reaction.
Illustrative Example
The reaction, 2N2O5(g)  4NO2(g) + O2(g), takes place in a closed
container, it was found that the concentration of NO2 increases by
1.6 x 10-2 moles/litre in four seconds. Calculate the rate of the
reaction and the rate of change of the concentration of N2O5.
Solution:
2N2O5(g)  4NO2(g) + O2(g)
rate of change of concentration of N2O5
d[N2O5 ]
1 d[N2O5 ]
or = 2×rate
2
dt
dt
1 d[NO2 ] 1 1.6×10-2
but rate =
= ×
=1.0×10-3 molL-1 s-1
4
dt
4
4
d[N2O5 ]
= 2×10-3 molesL-1 s-1
dt
=-
Rate law
It is a mathematical expression that gives the true
rate of reaction in terms of concentration of the
reactants which actually influence the rate.
Rate expression is experimentally determined.
Consider a general reaction

CH3 CHO(g)  CH4 (g)  CO(g)
dCH4 
dt
3/2
= k CH3CHO
Molecularity
For an elementry reaction, molecularity
is the total number of molecules taking
part in reaction.
It has no significance for complex reaction
Order of reaction
The number of molecules which actually react and undergo a
change in concentration is known as order of reaction.
For example,the dissociation of N2O5 is a first order reaction
2N2O5  4NO2  O2
There can be a reaction of fractional order
Consider a general reaction
aA +bB 
 cC + dD
The rate law
dx
a
b
= k A  B
dt
overall order = a + b
Molecularity vs. Order
Molecularity of Reaction
Order of Reaction
It is the number of atoms, ions or
molecules that must collide with
one another simultaneously so as
to result into a chemical
reaction.
It is the sum of the power of
concentration terms on which the
rate of reaction actually depends or
it is the sum of powers of the
concentration terms in the rate law
equation.
Molecularity of reaction
Cannot be zero.
Order of reaction can be zero.
It is a theoretical concept.
It is determined experimentally.
It is always a whole number.
It can even have fractional values.
The overall molecularity of
complex reaction has no
significance. Individual step has its
own molecularity.
Order of reaction is for overall
reaction.
Illustrative Example
For a reaction, aA + bB + cC  proucts, rate expression
can be expressed as rate = k[A]1[B]-3/2[c]1/2, Write the
order of the reaction with respect to ‘A’ , ‘B’, and ‘C’.
What is the overall order of the reaction?
Solution :
Order of the reaction w.r.t. ‘A’ = 1
Order of the reaction w.r.t. ‘B’ = -3/2
Order of the reaction w.r.t. ‘C’ = 1/2
Overall order of the reaction = 1 + (-3/2) + 1/2 = 0
Illustrative Example
The rate of the reaction, A  P, increases by
a factor of 5.18 when the concentration of A is
tripled. What is the order of A?
Solution:
Assuming the order of the reaction w.r.t. ‘A’ is n, then
the rate expression can be written as
rate(rA ) = -
d[A]
= k[A]n
dt
rA
n
 [A]1 
 1 =

rA
[A]2 

2
rA
n
 [A]1 
1
or
=

5.18rA
3[A]

1
1
or n = 1.5
Illustrative Problem
For the reaction
A+B
 Products
The following initial rates were obtained at various given intial
concentrations
S. No
[A]
[B]
Rate (mol lt–1 sec–1)
1.
0.1
0.1
0.05
2.
0.2
0.1
0.10
3.
0.1
0.2
0.05
Write rate law and find the rate constant of the above reaction.
Solution
-
dx
=k[A]'[B]o =k[A]
dt
Rate constant, k =
0.05
=5×10-1 s-1
0.1
Zero order reaction
A
Product
Rate of reaction
d[A]
0
= - k A 
dt
where k = rate constant for zero order reaction
 d[A] = - k
dt
[A] =- kt + C
When
where C = Integration Constant.
t = 0, then [A] =A0 hence C = A0
kt  [A]0  [A]
Zero order reaction
Half life of a reaction :- Time taken for the reactant
concentration to decrease to one half of its initial value
Ao
A  =
2
k t1/2 =  Ao  t1/2
Ao 

=
 Ao  = Ao
2
2
2k
Half life a initial concentration
Zero order reaction
Half life a initial concentration
t1 / 2
Initial concentration
Concentration [A]
Zero order reaction
t1/2
dx
dt
Time 't'
Initial conc. [A]0
t
Graphical representation for zero order reaction on the basis of x = kt + C
First order reaction
A
Product
initial conc. a
at time t
(a  x)
-
d[A]
= k[A]
dt
a-x
-
0
x

a
t
d[A]
d[A]
= k
dt
dt
0
ln[A]a
a-x = kt
ln
a
= kt
a- x
Unit of rate constant is time-1
(mol / lit)1n time1
n  order of reaction
k 
2.303
a
log10
t
(a  x)
First order reaction
kt
 loga
2.303
Plot of (a  x) vs t is a straight line
k
with slope 
2.303
log(a  x) 
When t  t1 / 2
t1 / 2 
0.693
k
Half life of first order reaction does not
depend upon initial concentration
Amount of substance left after
n half lives 
Ao
2n
First order reaction
For 75% completion of reaction
1
4
2.303
a
x
log
t75%
a/4
ax 

2.303
log 4
t75%
t75%  2t1 / 2
First order reaction
log[A]o
log [A]
t
Fig: Graphical representation for first order reaction
Slope of line = -
k
2.303
Illustrative Example
For a first order reaction, A  P, half life is
69.3 min. Calculate the time taken to decay
the amount of ‘A’ to 80%.
Solution:
t1 / 2  69.3 min or
t
0.693
t1 / 2
0.693
 0.01min1
69.3

k
k 
2.303
100
log
[ 80% reaction is complete]
t
20
2.303
2
10
log5  160.97 min.
Alternate Solution
n
1
We know, Nt  No  
2
n  no. of half life
n
1
20  100  
2
n

1 1

5  2 
log
n 
1
1
 nlog
5
2
0.699
 2.322
0.301
 time  2.322  69.3  160.9 min.
Illustrative Example
The half-life of a substrate in a certain enzyme-catalyzed
first order reaction is 138 s. How long will it require for
the initial concentration of substrate, which was 1.28
mmol/L, to fall to 0.040 mmol/L?
Solution:
t
[A]  2.303  138
2.303
 1.28 
log  0  
log 

k
0.693
 0.040 
 [A] 
t  690 s
Pseudo - first order reaction
Reactions which are not truly of the first order but
under certain conditions reactions become that of
first order are called pseudo unimolecular reaction.
For example: Hydrolysis of ester in presence of acid
CH3COOC2H5 + H2O  CH3COOH + C2H5OH
From this reaction, the rate expression should be
r = k [ester] [H2O]
Since, hydrolysis takes place in the excess of H2O and
concentration change of H2O is negligible practically.
therefore, r = k’ [ester]
Where k’ = k[H2O].
Thank you