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02 Chemical Kinetics

Introduction
Chemical Kinetics is the branch of physical chemistry which deals with the study of rate of reactions,
the mechanism by which the reactions proceed and factors affecting rate of reaction.
On the basis of rate, chemical reaction are broadly divided into three categories:
Chemical Reactions

Very fast Very Slow Moderate

Or reactions Or
Instantaneous Slow reactions
reactions

(A) Very fast or instantaneous reactions: Generally these reactions involve ionic species and
known as ionic reactions. These reactions take about 10-14 or 10-16 seconds for completion. So,
it is almost impossible to determine the rate of these reactions.
Examples:
AgNO3 + NaCl → AgCl  + NaNO3
(white ppt)
BaCl2 + H2SO4 → BaSO4  + 2HCl
(white ppt)
HCl + NaOH → NaCl + H2O
(B) Very slow reactions: These reactions proceed very slowly, may take days or months to show
any measurable change at room temperature.
Examples:
• Rusting of iron.
• Reaction between H2 and O2 to form H2O at ordinary temperature in absence of catalyst.
• CO + 2H2 ⎯⎯⎯⎯⎯⎯
at room temperature
→ CH3OH
(C) Moderate or slow reactions: This type of reactions proceed with measurable rates at normal
temperature and we can measure the rate of these reactions easily. Mostly these reactions are
molecular in nature.
Examples:
• Decomposition of H2O2
2H2O2 → 2H2O + O2
• Decomposition of N2O5
2N2O5 → 4NO2 + O2
• Hydrolysis of ester
CH3COOC2H5 + NaOH → CH3COONa + C2H5OH
• Inversion of cane sugar in aqueous solution
• Reaction of NO with chlorine
NO + Cl2 → NOCl2
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Rate of Reaction
Rate of reaction is defined as the change in concentration or pressure of reactant or product per unit
time. It is always a positive quantity.

Change in concentration of reactant of product

Rate of reaction =
Time taken in change

Where C = change in concentration in a small interval t

[+] sign is used when we refer for product concentration.

Concentration
[–] sign is used when we refer for reactant concentration. Product

P
For gaseous reactions r =  (unit of rate = pressure time–1)
t
Reactant
1  P  Time
and r = ×  (unit of rate = M time–1)
RT  t 
Types of Rate of Reactions:
Types of Rate of reactions

Average Rate of Reaction Instantaneous Rate of Reaction

(A) Average Rate of Reaction

The rate of reaction over a certain measurable period of time during the course of reaction is called
average rate of reaction. It is denoted by r .
For a reaction A → B
 [A]2 − [A]1  [A]
raverage =  =
 t 2 − t1  t
Where [A]1 = Concentration of reactant A at time t1,
[A]2 = Concentration of reactant A at time t2.
(B) Instantaneous Rate of Reaction
The rate of reaction at any particular instant during the course of
reaction is called instantaneous rate of reaction.
For a reaction A → B
[A] d[A]
Mathematically ; Instantaneous rate = lim (Average rate)
t →0

 [A]   [B]  dt 
rinst = lim  −  = lim   t(time)
t → 0
  t  t →0
 t 
𝐝ሾ𝐀ሿ 𝐝ሺ𝐁ሻ
or 𝐫𝐢𝐧𝐬𝐭 = ሺ−ሻ = ሺ+ሻ
𝐝𝐭 𝐝𝐭
Hence, Slope of the tangent at time t in plot of concentration with time gives instantaneous rate of
reaction.
 C  dC
Instantaneous rate of reaction = lim    =
t →0  t  dt
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Chemical Kinetics
Rate of Reaction in the Form of Stoichiometry of a Chemical Reaction
Let us consider a reaction: m1A + m2B → n1P + n2Q
d[A]
Where, Rate of disappearance of A = –
dt
d[B]
Rate of disappearance of B = –
dt
d[P]
Rate of appearance of P =
dt
d[Q]
Rate of appearance of Q =
dt
1  d[A]  1  d /[B]  1 d[P] 1 d[Q]
Rate of Reaction = − = − = =
m1  dt  m2  dt  n1 dt n2 dt
• Rate of reaction is always positive; negative sign represents decrease in concentration of reactant.
Units of Rate of Reaction
Unit of rate of reaction = mol L–1 time–1 i.e. (mol L–1 s–1 or mol L–1 min–1 or mol L–1 h–1)
Illustration 1:
For the reaction N2 + 3H2 ⎯⎯
→ 2NH3, if rate of appearance (ROA) of NH3 = 2×10–4 mol L–1 s–1
Then calculate rate of disappearance (ROD) of N2 & H2 and also calculate rate of the reaction ?
Solution:
N2 + 3H2 → 2NH3
1 1
ROR = 1 × ROD of N2 = × ROD of H2 = × ROA of NH3
3 2
1
ROD of N2 = × 2 × 10–4 = 10–4 ms–1
2
3
ROD of H2 = × 2 × 10–4 = 3 × 10–4 ms–1
2
1 1
ROR = × ROA of NH3 = × 2 × 10–4 = 10–4 ms–1
2 2
Illustration 2:
If the reaction 2NO2 → 2NO + O2, the rate of formation of NO is 6 g min–1, calculate rate of disappearance
of NO2 in g min–1.
Solution:
2NO2 → 2NO + O2
1 1
ROR = × ROD of NO2 = × ROF of NO = 1 × ROF of O2
2 2
2
ROD of NO2 = × ROF of NO
2
6 mol
ROD of NO2 = 1 × = 0.2 mol min–1 (molar mass of NO = 30)
30 min .
(ROD of NO2 )g min−1 = 0.2 × 46 = 9.2 g min–1 (molar mass of NO2 = 46)
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Illustration 3:
A gaseous hypothetical chemical equation 2A ⇌ B+C is carried out in a closed vessel. The concentration of B is
found to increase by 5 × 10–3 mol L–1 in 10 second. The rate of appearance of B is :
Solution:
Increase in concentration of B = 5 × 10–3 mol L–1 Time = 10 sec
increase of conc. B
Rate of appearance of B =
Time taken
5  10−3 molL−1
= = 5 × 10–4 mol L–1 sec–1
10 sec

BEGINNER’S BOX-1

1. The rate of a reaction is expressed as :

1  C 1  D 1    A     B 
+ = = −  =− 
2 t 3 t 4  t   t 
Then reaction is
(1) 4A + B → 2C + 3D (2) B + 3D → 4A + 2C (3) A + B → C + D (4) B + D → A + C
  A  B
2. In the reaction, A + 2B → 6C + 2D if − is 2.6 × 10–2 M s–1, what will be the value of − ?
t t
(1) 8.5 × 10–2 M s–1 (2) 2.6 × 10–2 M s–1 (3) 5.2 × 10–2 M s–1 (4) 7.5 × 10–2 M s–1
3. In the following reaction, how is the rate of appearance of the underlined product related to the
rate of disappearance of the underlined reactant
BrO3− ( aq ) + 5Br −1 ( aq ) + 6H+ ( aq ) → 3Br2 ( ) + 3H2O ( aq )

 BrO3−   Br2  1  BrO3   Br2 

−

(1) ( − )  = (2) ( − ) =
t t 3 t t
 BrO3−  1  Br2 
(3) ( − )  = (4) None of these
t 3 t

Rate Law
The experimental expression of rate of reaction in terms of concentration of reactants is known as rate law.
In this expression the rate of a reaction is proportional to the product of molar concentration of reactants
with each term raised to the power or exponent that has to be found experimentally.
In a chemical reaction: aA + bB → Product
The rate law is: Rate  [A]x [B]y
The values of exponents x and y are found experimentally which may or may not be same as stoichiometric
coefficients.
Above relationship can be written as :-
Rate = k[A]x[B]y
Where k is a proportionality constant known as rate constant.
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Chemical Kinetics
Rate constant:
In a chemical reaction –
n1A + n2B → m1C + m2D
according to law of mass action
Rate = k[A]n1 [B]n2
but according to rate law (experimental concept)
Rate = k[A]x[B]y
if [A] = [B] = 1 mol/L
then, Rate = k
Rate of reaction at unit concentration of reactants is called as rate constant or specific reaction rate. Rate
constant does not depend on concentration of reactant but it depends on temperature and catalyst.
Order of Reaction
The sum of powers of concentration of reactants in rate law expression is known as order of reaction.
For the reaction aA + bB → Product
Rate law is Rate = k[A]x[B]y
Here x = order of reaction with respect to A
y = order of reaction with respect to B
x + y = n (overall order of reaction)
Order of reaction may be zero, positive, negative or fractional.
Order of reaction is an experimental quantity.
Units of rate constant:
Rate = k[A]n
mol
 time−1
r unit of rate
k= = = L n
[A]n [unit of concentration]n  mol 
 L 
1− n
 mol 
k=   time−1
 L 
For gaseous reaction unit of k may be = (atm)1–n × time–1
Illustration 4:
For a reaction rate law equation is r = k(C)5/2. Then unit of rate constant will be:
Solution:
r = k(C)5/2
unit of r
Unit of K =
(unit of concentration)5/2
1 −5/2
 mol 
=  × sec–1 = mol–3/2 L3/2 s–1
 L 
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Illustration 5:
For a reaction the initial rate is given as: R 0 = k[A]02 [B]0 by what factor, the initial rate of reaction will increase
if initial concentration of A is 1.5 times and B is tripled?
(1) 4.5 (2) 2.25 (3) 6.75 (4) None of these
Solution:
R0 = K[A]02[B]0
[A]1 → [1.5A]0 , [B]1 → [3B]0
[B]1 → [3B]0
= K[1.5 A]02 [3B]0
= K × 6.75 [A]02 [B]0
 R1 = 6.75 R0
Molecularity
Total number of molecules, atoms or ions (reacting species) participating in an elementary reaction is
called as molecularity of reaction.
 Molecularity is a theoretical quantity.
 Molecularity can be an integer (1, 2 or 3) but it cannot be zero or negative or fractional.
 In elementary reaction molecularity is equal to its order.
 In complex reaction molecularity of each step of mechanism is defined separately.
 Total molecularity of complex reaction is meaningless.
 In complex reactions generally molecularity of slowest step is same as order of reaction which
can be considered as molecularity of reaction. (Except when slowest step contain intermediate)
 Maximum value of molecularity or order is 3 because chances of effective collision of more than
three molecules is very rare.
Mechanism of Reaction
(a) Elementary reactions:
Those reactions which completes in single step and which have exponents in rate law equal to
stoichiometric coefficients of the reactants.
If A + B → Products; is an elementary reaction
then rate law will be – Rate = k[A][B]
Zero order reactions can never be elementary reactions.
For elementary reactions fractional order is not possible.
Mechanism of Reaction
(b) Complex reactions:
Those reactions which complete in multisteps. For these reactions a mechanism is proposed.
 For complex reactions the overall rate of reaction is controlled by the slowest step which is
called as rate determining step (R.D.S.).
 In rate law expression rate of reaction depends on concentration of reactants of slowest step
which must be free from intermediate.
 If R.D.S. contains intermediate, its value is solved using Keq of fast step (assumed as reversible)
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Chemical Kinetics
Example-1
2NO2Cl → 2NO2 + Cl2
Experimentally, the rate law is Rate = k[NO2Cl]
The mechanism of the reaction is given as –
(i) NO2Cl → NO2 + Cl (slow step)
(ii) NO2Cl + Cl → NO2 + Cl2 (fast step)
So the rate law from slowest step Rate = k[NO2Cl]
In this way the predicted rate law derived from two step mechanism agrees with experimental
rate law.
Example-2
2NO(g) + 2H2(g) → N2(g) + 2H2O(g)
experimentally the rate law is, Rate = k [NO]2[H2]
The mechanism of the reaction is given as –
Kf
(i) 2NO Kb
N2O2 (fast step)
(ii) N2O2 + H2 → N2O + H2O (slow step)
(iii) N2O + H2 → N2 + H2O (fast step)
The Rate law from slowest step is:
Rate = k[N2O2] [H2]
The rate law expression should be free from intermediate species N2O2.
From fast reversible step –
kf[NO]2 = kb[N2 O2]
k
 [N2O2] = f [NO]2
kb
and rate law becomes
k 
Rate = k  f  [NO]2 [H2]
 kb 
Therefore, Rate = k'[NO]2[H2]
This derived rate law agrees with experimental rate law.
Pseudo First Order Reaction
A chemical reaction in which value of order of reaction is one but molecularity is more than one are known
as pseudo unimolecular/pseudo first order reaction.
Example-1
Hydrolysis of ester in acidic medium.
+
CH3COOC2H5 + H2O ⎯⎯
H
→ CH3COOH + C2H5OH
Rate = k [CH3COOC2H5][H2O]
Water is in excess then its concentration remain constant during the reaction and [H2O] is taken as
constant therefore,
Rate = k' [CH3COOC2H5] where k' = k [H2O]
Example-2
Inversion of cane sugar.
+
C12 H 22 O11 + H 2 O ⎯⎯
H
→ C6 H12 O6 + C6 H12 O6
Canesugar Glucose Fructose
Rate = k [C12H22O11][H2O]
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Water is in excess then its concentration remain constant during the reaction and [H2O] is taken as
constant therefore,
Rate = k' [C12H22O11] where k' = k [H2O]

 Reactant taken in excess can't affect order of reaction.

 In certain complex reaction product is also considered in order calculation.
 Order of reaction is determined experimentally.

Illustration 6:
For A + 2B → C
Rate law: r = k[A] [B]2; Find order of reaction if B is taken in excess
Solution:
For a reaction
A + 2B → C
Rate law; r = k[A][B]2
Since B is taken in excess
r = K[A]
(1st order reaction)

Half Life Time & Fractional Life Time

tfraction or tx: The time in which x fraction of initial amount of reactant is consumed; is called fractional
life.
a
if, x = ; t1/2 or t50%
2
3a
x= ; t3/4 or t75%
4
7a
x= ; t7/8 or t87.5%
8
x = a; tcompletion = t100%

BEGINNER’S BOX-2

1. A2 + B2 → 2AB; R.O.R. = k[A2] a[B2]b

[A2] [B2] Rate of reaction [Ms–1]
0.2 0.2 0.04
0.1 0.4 0.04
0.2 0.4 0.08
order of reaction with respect to A2 and B2 are respectively :
(1) a =1, b=1 (2) a=2, b=0 (3) a=2, b=1 (4) None
2. For a reaction the initial rate is given as : R0 = k[A]02[B]0 by what factor, the initial rate of reaction
will increase if initial concentration of A is 1.5 times and B is tripled ?
(1) 4.5 (2) 2.25 (3) 6.75 (4) None of these
3. For A(g) + B(g)→ C(g); rate = k[A] [B] , if initial concentration of A and B are increased by factor
1/2 2

of 4 and 2 respectively, then the initial rate is changed by the factor :-

(1) 4 (2) 6 (3) 8 (4) None of these
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Chemical Kinetics
Study of Different Order Reactions
Order of Reactions

Zero order First order th

n Order
Reactions Reactions Reactions

Second order Third order

Reaction Reaction
Zero order reactions
Reactions in which rate of reaction remains independent of concentration of the reactant are said to be
zero order reactions.
 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.
Example:
h
(a) H2(g) + Cl2(g) ⎯⎯ → 2HCl(g)
(b) 2NH3(g) ⎯⎯
Pt

→ N2(g) + 3H2(g)
(c) Reaction between Acetone and Bromine.
(d) Dissociation of HI on gold surface.
(e) Adsorption of gases on metal surface: At low P, rate of adsorption is proportional to surface area covered
which is proportional to P or concentration of gas hence order is 1 whereas at high P, complete surface
gets covered by gas & rate becomes independent of P & concentration hence order is 0.
Differential Rate Equation
A → Product
t=0 a 0
t=ts (a – x) x
d[A]
– = k[A]0
dt
dx
= k[A]0
dt
Calculation of Rate Constant
Let us take the reaction
A → Product
d[A]
– = k[A]0 = k
dt
–  d[A] =  kdt
– [A]t = kt + C
at t = 0 [A]t = [A]0 Slope = –k
[A]t
– [A]0 = k × 0 + C
C = – [A]0
On substituting the value of C t
– [A]t = kt – [A]0

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[A]t = [A]0 − kt [Integrated rate equation]
y = c – mx
[A]0 – [A]t = kt
[A]0 – [A]t x
x Slope =k
=
t t
x = kt

t
x = Amount of reactant that will change in product.
 For zero order reaction, rate of reaction is equal to rate constant

Unit of rate constant

k = mol L–1 s–1 = unit of rate of reaction.
Half-life period – The time in which half of the initial amount of reactant is consumed.
[A]0
At t = t1/2, [A]t =
2
[A]0 [A]0
 k t1/ 2 = [A]0 − or t 1/2 =
2 2k
The half life period for a zero order reaction is directly proportional to the initial concentration of
the reactants.
Time for completion of reaction
[A]t = [A]0 – kt
For competition [A]t = 0
[A]0 [A]0
k= t 100% =
t k
Graphical representation

Zero Order
t1/2
Rate

Concentration [A]0

Illustration 7:
For a reaction concentration of reactant is 0.5 M after 30 s. Then, find out initial concentration of reactant
(k = 10–2 mol L–1 s–1)
Solution:
Since unit of k = mol L–1s–1  Zero order reaction.
[A]t = [A]0 – kt
0.5 = [A]0 – 10–2 × 30
[A]0 = 0.5 + 0.3
[A]0 = 0.8 M
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Chemical Kinetics
Illustration 8:
In a zero order reaction, if time for 87.5% completion is 5 min then t50%?
Solution:
7 [A]0
t87.5 =
8 K
[A]0
t50% =
2K
t 87.5% 7
=
t 50% 4
4
t50% = t87.5%
7
4 20
t50% = × 5 = min
7 7

First order reactions

Reactions in which the rate of reaction is directly proportional to concentration of reactant.
Example :
(a) 2N2O5 ⎯⎯
→ 4NO2 + O2
(b) NH4NO2 ⎯⎯
→ N2 + 2H2O
+
(c) CH3COOC2H5 + H2O ⎯⎯
H
→ CH3COOH + C2H5OH
(d) 2H2O2 ⎯⎯
→ 2H2O + O2
(e) All radioactive decay

Differential rate equation

A → Product
t=0 a 0
t = t s (a – x) x
d[A] dx
– = k[A] = k (a − x )
dt dt

Calculation of rate constant

d A
– = k  dt
 A
–ln[A]t = kt + c ...(i)
At t = 0; [A]t = [A]0  C = – ln[A]0
Putting the value of C in equation (i)
–ln[A]t = kt – ln[A]0
ln[A]t = ln[A]0 − kt
y = c – mx

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 A 0  a 
ln = kt kt = ln   ...(ii)
 A t a−x
 A 0
2.303 log10 = kt
 A t
2.303 [A]
K= log 0
t [A]t
from equation(ii)
[A]o
= e kt
[A]t
 A t
= e− kt  [A]t = [A]0 e− kt Wilhemy equation
 A o
Unit of rate constant [k = time–1]
Half-life Period: The time in which half of the initial amount of reactant is consumed.
At t = t1/2 ; x = a/2 ; a–x = a/2
1  a  n2 2.303 0.693
t1/2 = n  or t1/2 = = (log2) or t 1/2 =
k  a /2  k k k
Half-life period for first order reaction is independent of the initial concentration of reactant.
Time for 3/4th of the Reaction (t3/4): The time in which 3/4th of the initial amount of reactant is consumed.
At t = t3/4 ; x = 3a/4 ; a–x = a/4
2.303 2.303
t3/4 = (log4) = × 2 log 2 = 2× t1/ 2
k k
t3/4 for first order reaction is independent of the initial concentration of reactant.
2.303 a − x1
Interval Formula k= log
t2 − t1 a − x2
where x1 and x2 are the amount consumed at time t1 and t2 respectively.
Time required for the completion of definite fraction of the first order reaction is independent of the initial
concentration of the reactant.
Graphical Representation

t1/2 Slope = k
Rate 2.303
log(a - x)

Concentration a t
r=k[A]

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Chemical Kinetics
Illustration 9:
If t1/2 of 1st order reaction is 30 min. then find ROR when concentration of reactant is 0.8 M?
Solution:
0.693
t1/2 =
K
0.693
30 =
K
0.693
K= min–1
30
0.693
r = K[A]1 = × 0.8 = 0.018 mol L–1min–1
30
Illustration 10:
In first order reaction, time for 90% completion is 15 min. then find time for t99.9% completion?
Solution:
2.303 100
K1 = log
t 90% 100 − 90
2.303 100
K2 = log
t 99.9% 100 − 99.9
K1 = K2 = same
2.303 2.303
log10 = log1000
t 90% t 99.9%
t99% = 3t90% = 3 × 15 = 45 min
Illustration 11:
How much time is required for conversion of reactant from 20 g to 0.625 g in 1st order reaction (t12 is 5 min)?
Solution:
20 ⎯⎯ 5
→ 10 ⎯⎯ 5
→ 5 ⎯⎯ 5
→ 2.5 ⎯⎯
5
→ 1.25 ⎯⎯ 5
→ 0.625
Total time = 25 min
[A]0 20
or [A]t = n
→ (2)n =
(2) 0.625
(2)n = 32 → n = 5
total time
5=
5
Total time = 25 min
First Order Kinetics in Terms of Volume

1  V − V0 
k= n 
t  V − Vt 
Where, V = volume at the end of the reaction.
Vt = volume at time t
V0 = Volume at time t = 0

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First Order Kinetics in Terms of Pressure
1  P − P0 
k= n 
t  P − Pt 
Where, P = Pressure at the end of the reaction.
Pt = Pressure at time t
P0 = Pressure at time t = 0
Illustration 12:
2A(g) → 2B(g) + C(g) is a first order reaction with initial pressure P0 and after time t pressure is Pt then
calculate K ?
Solution:
2A(g) ⎯⎯→ 2B(g) + C(g)
P0 0 0
(P0 – 2x) 2x x
Pt = P0 – 2x + 2x + x = P0 + x
 x = Pt – P0
2.303 P0 2.303 P0
K= log = log
t P0 − 2x t P0 − 2(Pt − P0 )
2.303 P0
K= log
t 3P0 − 2Pt
Illustration 13:
A(g) → 2B(g) + C(s) is a first order reaction with initial pressure P0 and after time t pressure is Pt then calculate K ?
Solution:
A(g) ⎯⎯→ 2B(g) + C(s)
P0 0 –
(P0 – x) 2x –
Pt = P0 – x + 2x = P0 + x
 x = Pt – P0
2.303 P 2.303 P0
K= log 0 = log
t P0 − x t 2P0 − Pt
2.303 P0
K= log
t 2P0 − Pt
General Integrated Rate Equation (nth order kinetics)
1  1 1 
kt =  n −1
− n −1  [n ≠ 1]
(n − 1)  (a − x) a 
a
If t = t1/2 ; x=
2

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Chemical Kinetics

1  2  1 
n −1 n −1

Therefore kt 1/2 =  −  
( n − 1)  a  a 

1  2n −1 − 1  ( n  1) ; t 1/2 
1
kt 1/2 =  
(n − 1)  a n −1  a n −1

Second Order Reactions

Hydrolysis of ester by alkali (Saponification)
CH3COOC2H5 + NaOH → CH3COONa + C2H5OH
H2 + I2 → 2HI 2HI → H2 + I2
2NO2 → 2NO + O2 2NO2 + F2 → 2NO2F
NO + O3 → NO2 + O2 2Cl2O → 2Cl2 + O2
1  1 1 1 1
For second order: n = 2 kt =  − = −
(2 − 1)  (a − x) a  (a − x) a
1
Half life t 1/2 =
ak

1 1
(a − x) k

Third Order Reactions

2NO + O2 → 2NO2 2NO + Cl2 → 2NOCl
2NO + H2 → N2O + H2O 2FeCl3 + SnCl2 → 2FeCl2 + SnCl4
1  1 1  1 1 1
for third order n = 3 kt =  − 2=  − 2
(3 − 1)  (a − x ) a  2 (a − x) a 
2 2

3
Half life t 1/2 =
2a2k

1 3
(a − x)2 2k

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Illustration 14:
The rate constant for a second order reaction is 8 × 10–5 M–1 min–1: How long will it take a 1M solution to
be reduced to 0.5 M.
Solution:
1 1
∵ = Kt +
[A]t [A]0
1 1
= (8 × 10–5)t +
0.5 1
1 = (8 × 10–5)t
1
t= = 1.25 × 104 min
8  10−5

BEGINNER’S BOX-3

1. Which of the following expressions is correct for zero order and first order reactions respectively
(where a is initial concentration) ?
1 1
(1) t 1/2  a;t 1/2  (2) t 1/2  a;t 1/2  a0 (3) t 1/2  a0 ;t 1/2  a (4) t 1/2  a;t 1/2 
a a2
2. For the zero order reaction, A →B + C; initial concentration of A is 0.1 M. If [A] = 0.08 M after 10
minutes, then it's half-life and completion time are respectively :
(1) 10 min; 20 min (2) 2 × 10–3 min; 4 × 10–3 min
(3) 25 min, 50 min (4) 250 min, 500 min
3. For an elementary reaction, X(g) → Y(g) + Z(g) the half life period is 10 min. In what period of
time the concentration of X will be reduced to 10% of original concentration?
(1) 20 min (2) 33 min (3) 15 min (4) 25 min
4. A first order reaction is 75% completed in 100 min. How long will it take for it's 87.5%
completion?
(1) 125 min (2) 150 min (3) 175 min (4) 200 min
5. The rate constant for a first order reaction which has half life 480 s is :-
(1) 1.44 × 10–3 s–1 (2) 1.44 × s–1 (3) 0.72 × 10–3 s–1 (4) 2.88 × 10–3 s–1

Collision Theory of Chemical Reactions

This theory was given by Max Trautz and William Lewis. According to it, for a reaction to occur there must
be collisions in between reacting molecules. Total number of collisions per second in unit volume is called
collision frequency(z). Generally its value is very high for gaseous reactions (1025 to 1028 collisions/sec-cm3).
But only a small fraction of these collisions are capable to convert reactant into product. These collisions
are called as effective collisions.
For effective collision following two conditions must be satisfied at a time:
(a) Reacting molecules must posses a minimum amount of energy.
(b) Proper orientation of collision
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Chemical Kinetics
• Threshold energy: The minimum energy which must be possessed by reacting molecules for a
chemical reaction to occur.
• Activation energy: The minimum extra amount of energy required by reactant molecules for
converting into products.

Activated complex Activated complex

H=+ve Threshold
Threshold

Ea(b)
Ea(f)

Ea(f)
energy

energy
Ea(b)

Energy
Energy

P
R
H=–ve

R
Progress of Reaction Progress of reaction
H = E a (f) − E a (b ) H = E − E
Exothermic reaction Endothermic reaction
a (f) a (b)

(Ea)f < (Ea)b ; H = –ve (Ea)f > (Ea)b ; H = +ve

1. Ea(f) = Activation energy for forward reaction
2. Ea(b) = Activation energy for backward reaction
If not specified in questions then consider Ea for forward reaction.
H = E a ( f ) − E a ( b ) ; H = H p − H R

Activation Energy Mainly Depends Upon:

(i) Nature of reactant: For different reactants, number of bonds and bond energies are different,
therefore activation energy will also be different.
Reactions which have less Ea, take place at faster rate.
(ii) Presence of catalyst: Catalyst provide an alternative path of reaction mechanism for the reaction.
• In presence of catalyst threshold energy decreases, activation energy decreases and rate of
reaction increases.
• In presence of negative catalyst (inhibitor) threshold energy increases, activation energy
increases, rate of reaction decreases.

T.E.’’
Inhibitor
T.E.
absence of catalyst
Energy →

T.E.’
E’’a Ea Catalyst
E’a

Progress of Reaction
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(b) Orientation:

 Limitations:
(i) This theory is mainly applicable for gaseous reactions and also for solutions in which reacting
species are molecules.
(ii) This theory is mainly applicable for simple bimolecular reactions but fails for complex reactions.
(iii) It considers molecules to be hard sphere and ignore structural aspect of molecules.

• Activated complex is most unstable complex formed in the transition state with effective collision.
• On increasing temperature Ea for reaction does not decrease but number of active molecules which
are crossing the energy barrier increases therefore rate of reaction increases.
• According to Arrhenius ; rate of reaction  e–Ea/RT Ea = Activation energy
R = Gas constant
T = Temperature (in K)

Illustration 15:
The E a for an exothermic reaction A → B is 80 kJ mol –1. Heat of reaction is 20 kJ mol –1 . Ea for the
reaction B → A will be ?
Solution:
H = Ea(f) –Ea(b)  –20 = 80 –Ea(b)
Ea(b) = 100 kJ mol–1
Illustration 16:
For the reaction A + B C + D the activation energy is 32 kJ mol–1. For the reverse reaction the Ea is
58 kJ mol–1. Determine
(i) Nature of reaction (ii) H
Solution:
H = Ea(f) –Ea(b)
H = 32 – 58
H = –26 kJ mol–1 (exothermic)
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Chemical Kinetics
Factors Affecting Rate of Reaction

Nature of reactant
1 6 Presence of light

Concentration of 5 Catalyst
reactant 2

Pressure (for gases) 4

3 Temperature

1. Nature of reactant:
(a) Physical state of reactant:
increasing order of rate of reaction – Solid < liquid < gas
(Intermolecular attractive force decreases which provides more freedom for collisions)
(b) Physical size of particles (if reactant is solid):

1
Rate of reaction   surface area
physical size

(c) Chemical nature of reactant: For different reacting species number of bonds broken and their
bond energies are different. Therefore requirement of activation energy is also different. Now
reactions having less value of activation energy will take place at faster rate.
2. Concentration of reactant: Rate of reaction  concentration of reactant
3. Pressure: Effect of pressure on Rate of reaction is negligible when reactants are solid or liquid. But
if reactants are in gaseous state then rate of reaction increases on increasing pressure because
number of effective collisions increases.
4. Temperature: On increasing temperature rate of reaction increases whether the reaction is
exothermic or endothermic. When temperature increases KE of molecules increases, number of
activated molecules increases thus rate of reaction increases.
Relation between rate constant and Temperature:
(a) Generally it is found that for every 10 °C rise in temperature Rate of reaction becomes 2 to 3 times.
Temperature coefficient (): It is defined as ratio of rate constant of a reaction at two
different temperatures which will be differ by 10 °C.

k T+10 r k
= = 2to3 ; 2 = 2 =  T/10
kT r1 k 1

If temperature of reaction is not specified then consider 25 °C.

(If  is not given consider it as minimum 2)
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Maxwell And Boltzmann Energy Distribution Curve

T1

This area shows

Energy of activation Fraction of additional
Fraction of molecules

Molecules which
react at (T + 10)

This area shows

Fraction of molecules
Reacting at T

Kinetic energy
Illustration 17:
If temperature of a reaction is increased from t1 to t2 then rate of reaction becomes?
Solution:

()t2 −t1 /10  ()T/10 times ; rnew = rold × ()T/10

Illustration 18:
If temperature of a reaction is increased from 10°C to 90°C then how many times rate of reaction will
become?
Solution:
28 times
Illustration 19:
A reaction is carried out at 10°C. If temperature is increased by 50°C then how many times rate of reaction
will become?
Solution:
32 times
(b) Arrhenius equation
− Ea
K = Ae RT ...(1)

A = Arrhenius constant / pre-exponential factor / Frequency factor

E = Activation energy
R = gas constant
T = Temperature (kelvin)

 k increases with increase in temperature

If T →  ; k = A

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Chemical Kinetics
k
 = e− Ea / RT = fraction of molecules having energy  Ea
A
Ea
e−Ea /RT = Boltzman factor slope = −
Log10K 2.303R
On taking logarithm for equation (1) on both sides.
ln k = ln A + ln e− Ea / RT
Ea 1/T
2.303 log10k = 2.303 log10A –
RT
Ea
log10 k = log10 A − ...(2)
2.303RT
y = c – mx
 Ea of reaction can be determined by measuring rate constant at two different temperatures
At temperature T1:
Ea
log10k1 = log10 A – ...(3)
2.303RT1
At temperature T2
Ea
log10k2 = log10A – ...(4)
2.303RT2
Equation (4) – Equation (3) gives -
Ea  1 1 
logk2 – log k1 =  − 
2.303R  T1 T2 

k2 Ea  T2 − T1 
log10 =  
k1 2.303R  T1T2 

 From arrhenius equation - k = Ae− Ea /RT

Ea
ℓ nk = ℓ nA –
ER
d d d  Ea 
( nk) = ( n A) + −
dT dT dT  RT 
Ea d ( −1 ) Ea ( −2 )
= 0− T = T
R dT R
d E
nk = a 2 differential form of Arrhenius equation
dT RT
5. Presence of catalyst: In presence of catalyst Ea of reaction decreases and rate of reaction
increases.
6. Exposure to radiation: Rate of some reactions also increases when reaction are carried out in the
presence of radiation. (only for photochemical reaction)
e.g. formation of HCl
H2 + Cl2 ⎯⎯
→ 2HCl (very slow reaction)
H2 + Cl2 ⎯⎯
h
→ 2HCl (explosive)
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Illustration 20:
For a reaction, temperature coefficient = 2, then calculate the activation energy (in kJ) of the reaction.
Solution:
Let T1 = 25 °C , T2 = 35 °C
k2 Ea  T2 − T1 
log =  
k 1 2.303R  T1T2 
k2
Given: Temperature coefficient = =2
k1
T1 = 25 + 273 = 298 k,
T2 = 35 + 273 = 308 k, R = 8.314
Ea  10 
log 2 = 
2.303  8.314  298  308 
Ea = 52.897 kJ
Illustration 21:
An exothermic reaction A → B has an activation energy of 17 kJ mol–1 of A. The heat of the reaction is 40 kJ.
Calculate the activation energy for the reverse reaction B → A.
Solution:
For the reaction A → B.
Activation energy Ea = 17 kJ
H = – 40 kJ
H = Ea(f) – Ea(b)
Ea(b) = 17 – (–40)
= 57 kJ
Illustration 22:
For first order gaseous reaction log k when plotted against 1/T, gives a straight line with a slope of –8000.
Calculate the activation energy of the reaction.
Solution:
Arrhenius equation k = Ae–Ea/RT
Ea 1
log k = log A − 
2.303R T
1 Ea
when curve is plotted between log k and , a straight line is obtained. Slope of this line = −
T 2.303R
Ea
Then, = 8000
2.303R
or Ea = 8000 × 2.303 × 1.987
= 36608 Cal

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Chemical Kinetics
Parallel First Order Reaction
Reactant A converts into products B & C through parallel reactions of first order as following:
Keff = K1 + K2 B
K1 A
% Yield of B =  100
K1 + K 2
C

BEGINNER’S BOX-4
1. For an endothermic reaction where H represents the enthalpy of reaction in kJ mol–1, the
minimum value for the energy of activation will be
(1) Less than H (2) More than H (3) Equal to H (4) Zero
2. The activation energy of the reaction, A + B → C + D + 38 kcal is 20 kcal, what would be the
activation energy of the reaction, C + D → A + B
(1) 20 kcal (2) – 20 kcal (3) 18 kcal (4) 58 kcal
k 350
3.  1 , this means that
k 340
(1) Rate increases with the rise in temperature
(2) Rate decreases with rise in temperature
(3) Rate does not change with rise in temperature
(4) None of the above
4. The plot of n k versus 1/T is linear with slope of
Ea Ea Ea Ea
(1) − (2) (3) (4) −
R R 2.303R 2.303R

BEGINNER’S BOX ANSWER KEY

Que. 1 2 3
BEGINNER'S BOX-1
Ans. 1 3 3

Que. 1 2 3
BEGINNER'S BOX-2
Ans. 1 3 3

Que. 1 2 3 4 5
BEGINNER'S BOX-3
Ans. 2 3 2 2 1

Que. 1 2 3 4
BEGINNER'S BOX-4
Ans. 2 4 1 1
ss

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