Chemistry module II physical chemistry II for IIT JEE main and advanced ranveer singh

12mm

Fullyed
Solv

Fullyed
Solv

Singh


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About the Author
Ranveer Singh has an M.Sc. in Chemistry and has been mentoring and teaching Chemistry to JEE aspirants for more
than 15 years. He aims for perfection and has an enormous passion when it comes to applying new methods to create
solutions. These are the qualities that make him stand out from the crowd as a teacher and instructor.

IIT JEE PC-V2_Prelims.indd 2

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Ranveer Singh

McGraw Hill Education (India) Private Limited
CHENNAI
McGraw Hill Education Offices

Chennai New York St Louis San Francisco Auckland Bogotá Caracas
Kuala Lumpur Lisbon London Madrid Mexico City Milan Montreal
San Juan Santiago Singapore Sydney Tokyo Toronto

IIT JEE PC-V2_Prelims.indd 3

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McGraw Hill Education (India) Private Limited
Published by McGraw Hill Education (India) Private Limited
444/1, Sri Ekambara Naicker Industrial Estate, Alapakkam, Porur, Chennai - 600 116

Physical Chemistry for JEE (Main & Advanced) Module-II
Copyright © 2017, McGraw Hill Education (India) Private Limited.
No part of this publication may be reproduced or distributed in any form or by any means, electronic, mechanical, photocopying, recording,
or otherwise or stored in a database or retrieval system without the prior written permission of the publishers. The program listings (if any)
may be entered, stored and executed in a computer system, but they may not be reproduced for publication.
This edition can be exported from India only by the publishers,
McGraw Hill Education (India) Private Limited
ISBN (13): 978-93-5260-603-0
ISBN (10): 93-5260-603-5
Information contained in this work has been obtained by McGraw Hill Education (India), from sources believed to be reliable. However, neither McGraw Hill
Education (India) nor its authors guarantee the accuracy or completeness of any information published herein, and neither McGraw Hill Education (India)
nor its authors shall be responsible for any errors, omissions, or damages arising out of use of this information. This work is published with the understanding
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Typeset at The Composers, 260, C.A. Apt., Paschim Vihar, New Delhi 110 063 and text and cover printed at

Cover Designer: Creative Designer
visit us at: www.mheducation.co.in

IIT JEE PC-V2_Prelims.indd 4

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Dedicated
to
Shri Amar Chand Ji Maharaj

IIT JEE PC-V2_Prelims.indd 5

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Preface

Chemistry forms an important part of all entrance tests. In my teaching career, spanning over a decade and half, I have felt that a
chemistry book based on the changing perceptions, needs, feedback and the experiences of the students and educators is needed
by the engineering aspirants. It is with this vision that the present book has been written.
This book is not a textbook. It is a refresher text to help students revise their lessons in the quickest possible way and in the
most effective manner. It does not over emphasise theories, as has been done in several other competitive books available in the
market. However, every care has been taken to ensure that no important theory is left out. This book has several new features:

coverage of the syllabus of JEE (Main + Advanced); a great number of solved numerical examples to acquaint students with
the application of several theories, solution at the end of each exercise and two levels of questions at the end of the chapters to
give readers an opportunity to assess their understanding of the chapters. The use of easily understandable language is at the
core of the author’s efforts.
The exercises given at the end of every chapter is further categorised into three difficulty level of questions and their patterns
asked in JEE along with the previous year questions with solutions.
● Level-I has the questions mainly suitable for JEE Main exam.
● Level-II contains slightly difficult questions suitable for JEE Advanced.
● Level-III has the highest questions of various patterns asked in JEEAdvanced (such as more than one correct answer,
comprehension, match the column and single digit integer).
I hope this book will help in motivating and encouraging the students towards the preparation forthe Chemistry portion of
the examinations. Every care has been taken to make the book error-free. However, some mistakes may have been crept in inadvertently. Constructive suggestions and comments from students and teachers would encourage me to make the book more authentic and acceptable in the next edition. We wish our young readers a great success at the engineering entrance examinations.
Ranveer Singh

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Acknowledgements

I express my gratitude towards the publisher and the members of the editorial team. I would be failing in my duty if I don’t
express my thankfulness to Mr. Sanjay Agrawal for encouraging me to start writing this book and helping me step by step while
preparing the manuscript.

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Contents

Preface

vii

Acknowledgements

ix

1.

2.

3.

Chemical Kinetics
● Key Concepts
1.1
● Solved Examples
1.14
● Exercise
1.17
■
Level I 1.17
■
Level II 1.23
■
Level III 1.26
● Previous Years’ Questions of JEE (Main & Advanced)

● Answer Key
1.35
● Hints and Solutions
1.36
Electrochemistry
● Key Concepts
2.1
● Solved Examples
2.9
● Exercise
2.15
■
Level I 2.15
■
Level II 2.18
■
Level III 2.22
● Previous Years’ Questions of JEE (Main & Advanced)
● Answer Key
2.29
● Hints and Solutions
2.30
Solution and Colligative Properties
● Key Concepts
3.1
● Solved Examples
3.9
● Exercise
3.16
■

Level I 3.16
■
Level II 3.19
■
Level III 3.20
● Previous Years’ Questions of JEE (Main & Advanced)
● Answer Key
3.26
● Hints and Solutions
3.27

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1.1–1.52

1.33

2.1–2.52

2.28

3.1–3.39

3.26

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x

4.

5.

6.

7.

Physical Chemistry-II for JEE (Main & Advanced)

Solid State
● Key Concepts
4.1
● Solved Examples
4.10
● Exercise
4.11
■
Level I 4.11
■
Level II 4.14
■
Level III 4.15
● Previous Years’ Questions of JEE (Main & Advanced)
● Answer Key
4.20
● Hints and Solutions
4.20
Surface Chemistry
● Key Concepts

5.1
● Solved Examples
5.8
● Exercise
5.9
■
Level I 5.9
■
Level II 5.11
■
Level III 5.15
● Previous Years’ Questions of JEE (Main & Advanced)
● Answer Key
5.19
● Hints and Solutions
5.20
Redox
● Key Concepts
6.1
● Solved Examples
6.10
● Exercise
6.15
■
Level I 6.15
■
Level II 6.18
■
Level III 6.19
● Previous Years’ Questions of JEE (Main & Advanced)

● Answer Key
6.27
● Hints and Solutions
6.28
Thermodynamics and Thermochemistry
● Key Concepts
7.1
● Solved Examples
7.19
● Exercise
7.24
■
Level I 7.24
■
Level II 7.28
■
Level III 7.30
● Previous Years’ Questions of JEE (Main & Advanced)
● Answer Key
7.38
● Hints and Solutions
7.40

IIT JEE PC-V2_Prelims.indd 10

4.1–4.31

4.18

5.1–5.24

5.18

6.1–6.44

6.26

7.1–7.58

7.37

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1

Chapter

Chemical
Kinetics

Key Concepts
INTRODUCTION TO KINETICS

Moderate
2H2O2 h 2H2O + O2

Thermodynamics deals with the feasibility of the reaction but
kinetic deals with the rate of the reaction.
Thermodynamics deals with the initial and final state of

the reactants and products but kinetics deal with the path by
which reaction is taking place.
P.E.

? (kinetics)

[R]

Simple reaction
Those chemical reactions which
takes place in a single step are
known as simple reaction.
aA + bB Ỉ P

Complex reaction
Simple reaction
DH

R

[P]

P
time

a

P.E.

R

r = K[A] [B]

CLASSIFICATION OF
CHEMICAL REACTION

P

b

Reaction coordinate

Complex
Those chemical reactions which take in more than one step.
P.E.

Very fast

Very Slow

NaCl + AgNO3 Ỉ AgCl Ø+ NaNO3

1
25∞C
Ỉ FeO
Fe + O 2 ỉỉỉ
2

HCl + NaOH Ỉ NaCl + H2O

I*

[O]
Cellulose ỉỉỉ
Ỉ CO 2
25∞C

Conc.
[P]

R

+ H 2O

Conc.

Time

r = k[A]m [B]n
(m, n) π (a, b)
Rate of reaction

[P]
Time

IIT JEE PC-V2_01.indd 1

Reaction coordinate

aA + bB Ỉ P

[R]

[R]

P

(1) Average Rate:

DC
Dt

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1.2

Physical Chemistry-II for JEE (Main & Advanced)
Conc.
[P]

[P] t2
[R] t1
[P] t1
[R] t2

98°F

[R]
t1

t2

O

–Ve sign indicate decrease in concentration of reactant

temperature
(Biological Enzyme)

At T Æ ; maximum rate
For majority of the reaction; on increasing temperature
by 10°C rate of the reaction increases two to three times.
(4) Catalyst

moles
or M/sec–1
- sec

(–) ve catalyst

P.E.

Without catalyst
(+) ve catalyst

Conc.
R

C

P

O

Reaction

R
O

O

temperature (K)

Time

È[ R]t2 - [ R]t1 ˘ [ P]t2 - [ P]t1
DC
= -Í
˙=
Dt
t2 - t1
Ỵ t2 - t1 ˚

Units of reaction rate:

Rate

Rate

t1

T

Catalyst increases the rate of reaction
(5) Solvent

Time

OC
Ê dc ˆ
=
ÁË ˜¯
dt t = t1 OT

Solvent
Polar

In general, rate of reaction changes with time
Conc.

O¢

[P]
O

Protic
SN1
H2 O
EtOH

Conc.

t1

Time

T¢

Time

CH3
H C N
(DMF)
CH3
O

(6) Surface area
For reaction involving solid surface area,
rate μ surface Area
(7) Radiation

C

CO¢
Ê dc ˆ
=
ÁË ˜¯
dt t = t1 OÂT Â

25C

ặ CH3 Cl + H Cl
CH4 + Cl2 æææ
hv

Factors affecting rate of the reaction:
(1) Nature of reactants
700∞C

C (graphite) + O 2 ỉỉỉ
Ỉ CO 2
K
1

25∞C

Ỉ NO reaction
CH4 + Cl2 æææ
Dark
K1 > K 2 \ r1 > r2

700∞C

C (diamond) + O 2 ỉỉỉ
Ỉ CO 2
K
2

(2) Concentration or volume or pressure
n

P
[ A] = A = A
aA(g) + bB(g) Ỉ P
V
RT
m
n
Rate = K [ A] [ B]
K
=
( PA ) m ( PB ) n
( RT ) m + n
(3) Temperature
Rate of the reaction generally increases with the increases in temperature.

IIT JEE PC-V2_01.indd 2

Aprotic
SN2

Non-Polar
C6H6
n-C6H14

Rate Constant (K)/Velocity constant/Specific reaction
rate
Rate constant is defined as the rate of the reaction. When concentration of each reactant is unity.
aA + bB h P
If
and

rate = K[A]m [B]n
[A] = 1 M
[B] = 1 M
then rate = K

Characteristic of Rate constant (K)
Rate constant for any particular reaction is constant at constant temperature. It does not change with concentration volume, pressure, time, etc.

5/13/2017 10:25:44 AM


Chemical Kinetics

Catalyst increases rate constant ‘k’.
K depends on temperature.
K = Ae

Ea
- RT

K
A

Time

KT +10

= 2 to 3
KT

Higher the value of rate constant; more will be the rate of
the reaction
Higher value of rate constant; suggest reactant is kinetically unstable.
Graphite at 700°C is kinetically unstable than diamond
Units of rate const. (K)
RhP
Rate = K[R]m Order of reaction(m)
Temperature Coefficient (m) =

Moles
Ê mole ˆ
= KÁ
˜¯
Ë
- sec
Ê moles ˆ
K =Á
˜¯
Ë

1- m

◊

m

1
M 1- m
=
sec

sec

Rate Law
aA + bB h cC + dD
1 Ê d [ A] ˆ 1 Ê d [ B ] ˆ 1 Ê d [C ] ˆ
Á˜ = Á˜ = Á+
˜
a Ë dt ¯ b Ë dt ¯ c Ë dt ¯
1 Ê d [ B]ˆ
m
n
= Á+
˜ = k[ A] [ B]
d Ë dt ¯

Rate =

d [ A]
Ỉ rate of disappearance of A.
dt
d [C ]
+
Ỉ rate of formation of C.
dt
m Ỉ order w.r.t. A.
n Ỉ order w.r.t. B.
m + n Ỉ overall order of the reaction.
-

Relation between different rate constants

d [ A]
= K ◊ a[ A]m [ B]m = K A [ A]m [ B]n
dt
d [ B]
= K ◊ b[ A]m [ B]m = K B [ A]m [ B]n
dt
d [C ]
+
= K ◊ c[ A]m [ B]m = KC [ A]m [ B]n
dt
d [ D]
+
= K ◊ d [ A]m [ B]m = K D [ A]m [ B]n
dt
K
K
K
K
K= A= B = C = D
a
b
c
d

IIT JEE PC-V2_01.indd 3

1.3

Molecularity
For a single step reaction: Molecularity is defined as number

of reactant molecules participating in balanced chemical reaction. Molecularity of reaction is generally 1 or 2. For few
reaction it is also observed to be 3.
Molecularity cannot be more than 3 because probability
of more than 3 molecules colliding at the same instance is
almost zero.
Molecularity cannot be –ve, zero, or fraction.
For complex reaction : each step has its own molecularity
RhP
If mechanism
R h P* + Q*
P* + Q* h I*
I* h P
but overall molecularity has no significance.
For elementary reaction; molecularity and order are same.
Molecularity of a single step reaction can be obtained by
reaction stoichiometry.
Example (1) A h B
or
A h B + C Molecularity = 1
Example (2) 2A h P
or
A+BhP
or
A h 1/2 P Molecularity =2
Example (3) 3A h P
or
A h 1/3 P
or
2A + B h P Molecularity = 3
Unimolecular:

N2O5 h N2O3 + O2
Bimolecular:
H2 + Cl2 h 2HCl
Tri/ter molecular: 2H2 + O2 h 2H2O
Order of the reaction
In the expression of rate law, order is defined as (coefficient)
of active mass.
aA + bB h P
Rate = K[A]m [B]n
m = Order w.r.t. A
n = Order w.r.t. B
(m + n) = overall order of reaction
Order of any reaction can be obtained only by experimental methods.
Order of reaction may have –ve, +ve, zero or fractional
value.
Order of reaction is related with reaction mechanism.
reaction mechanism may change with experiment condition order may also change with reaction condition:
R X + Nu(–)
(2°)

SN1
–
H2O R Nu + X \ Rate = K[R–X]
SN2 R Nu + X– \ Rate = K[R–X] [Nu–]
DMF

5/13/2017 10:25:45 AM


1.4

Physical Chemistry-II for JEE (Main & Advanced)

Zero order reaction
A
Ỉ P
t=0
a
—
t
(a – x)
x
- d [ A] + d [ P]
Rate =
=
= K [ A]∞
dt
dt
- d [ A]
=K
dt
a- x

Ú
0

[or]
f

- d [ A] = K Ú dt

(i)

+ d [ P]
=K
dt
x

t

0

0

Ú dx = K Ú dt

0

(a – x) – a = –Kt
(a – x) = –Kt + a
[A]t = –Kt + [A]0

x = Kt

(ii)

n -1

(iv)
(v)

a
= 2 ¥ t1/2 or 2 ¥ t50%
K
Few graphs related to zero order reaction
Conc.
[A]t

Conc.
[P]

10 Ê 50 ˆ
1- n
=Á
˜ =2=2
5 Ë 100 ¯
fi n = 0.
(3) Zero order reaction is 100% completed.
(4) For zero order reaction in the same time interval, equal
amount of reactants are consumed and concentration
of reactants at the same time intervals are in A.P. with
C.D. –Kt.
AhP
t=0
a
t
a – Kt
2t
a – K(2t)
3t

a – K(3t)
x = Kt
Examples:
(1) Photochemical reaction

[A]0

hv

Time

Time
Rate

P

K
Time

lnt1/2

t1/2

Slop = ½ K

C2H2 + H2 ỉỉỈ C2H6
First order reaction
A
h
P

t=0
aM
—
t
(a – x)
x
- d [ A] + d [ P]
Rate =
=
= K [ A]1
dt
dt
a- x

Ú

45°

a

ln(1/2K)
a

H2 + Cl2 ỉỉỈ 2HCl
(2) Reaction taking place at the solid catalyst surface.
Ni

K
t50% t100% Time

lna

Characteristics of zero order reaction
(1) Rate is independent of concentration and does not
changes as the reaction progress.
K [ A]
Rate =
[ A] + K ¢

IIT JEE PC-V2_01.indd 4

n -1

fi

t100% =

–K

(t1/2 ) I Ê aII ˆ
=
(t1/2 ) II ÁË aI ˜¯

(iii)

Half-life of zero order reaction
x = Kt
at t = t½ = t50% = when x = a/2
a
t1/2 =

2K
Ê 1 ˆ
ln t1/2 = ln a + ln Á
Ë 2 K ˜¯
Time of completion for zero order reaction.
x = Kt

Conc.

K [ A]
[ A]
(2) Half-life of zero order reaction
A
Ỉ P
t=0
100
–
t = 10 min
50
50
t = 15 min
25
75
t = 17.5 min 12.5
87.5
Form the above information, reaction is zero order reaction.
1
We know t1/2 μ n-1
a
If [A] >> K¢ then, Rate =

fi
fi

t

- d [ A]
= K Ú dt
[ A]
0

ln(a – x) – ln a = –Kt
ln(a – x) = –Kt + ln a
a-x
ln
= - Kt fi a - x = a e - Kt
a
a
ln
= Kt
a-x
K=

(i)

2.303
a
log10
t
a- x

(ii)

(iii)
(iv)

5/13/2017 10:25:46 AM


Chemical Kinetics

Few graph related to first order reaction
Conc. [A]0

Conc. a = [A]
0
[P]t

Half-life of first order reaction
2.303
a
log
K=
t
a-x
2.303
a
2.303
fiK=
K=

log
log 2
t1/2
a - a /2
t1/2

[P]

[A]
Time

Time

fi

t1/2 =

(a – x) = ae–kt
x = a(1 – e–kt)

0.693 ln 2
=
K
K

or t1/2 μ

1
1-1

a

t1/2

Degree of dissociation of first order reaction
A
h
P
a
a
—
1.0
a–x
a(1 – x/a)
a=

1.5

0.693/K

a

x
= 1 - e - kt
a

t

t=0
t = 10 min

t = 20 min
t = 30 min

ln[A]
Slope = –K
ln[A]0

time
log[A]t
Slope = –K/2.303
log[A]0

time
ln a/a-x
Slope = K

time
log a/a-x
Slope = K/2.303

time
Rate

h

A
100
50
25
12.5

P
—
50
75
87.5

Characteristic of first order reaction
(1) In first order reaction, in equal time interval, same fraction of reactants are consumed and concentration of reactants at same time interval are in GP with a common
ratio e–Kt.
A
h
P
t=0
a
t
ae–Kt
2t
ae–K(2t)
3t
ae–K(3t)
Since [A]t = [A]0 e–Kt
(2) Rate μ [A]t
0.693
K
a
Examples of first order reaction
(3) t1/2 μ

(1)

(2)
(3)
(4)

1

1-1

and t1/2 =

H+

CH3COOEt + H2O ỉỉỈ CH3COOH + EtOH
2H2O2 h 2H2O + O2
All nuclear reaction
Decomposition of N2O5

Second order reaction
Case I: When concentration of both reactants are same
A
h
P
t=0
a
–
t=1
a–x
x
x

or
time
–[Kt]

Rate = K[A]t = K[A]0 e

IIT JEE PC-V2_01.indd 5

dx
1 È
1
1 ˘
(a - x) - n =
dt Ú0
(n - 1) ÍỴ (a - x) n -1 a n -1 ˙˚
dx
= K (a - x) 2
dt

So

x

t

0

0

-2

Ú (a - x) dx = Ú K dt

5/13/2017 10:25:48 AM


1.6

fi

Physical Chemistry-II for JEE (Main & Advanced)

Kt =

Eq. (2) – (1) = Eq. (3) – (2)

1
1
1
1
- =
a - x a [ A]t [ A]0

= Kt =

Half-life of second order reaction
1
t1/2 =
K ◊a

Examples of second order reaction

(1) Hydrolysis of ester in basic medium
(2) Decomposition of actatdehyde
CH3CHO h CH4 + CO

Graphical representation of second order reaction
Conc.

Case II: When concentration of both reactants are
different
A
+
B
Ỉ
P
t=0 a
b
–
t=t
a–x
b–x
x

Rate = K[A]2

Rate

P

[A]
Time

1
1
[ A]2t [ A]t

dx
= K (a - x)(b - x)
dt

Time

x

Slope = K
t1/2

x

1/[A]0

ln t1/2

t

1/a

1 1
Since t1/2 = K ¥ a

fi

x
x
˘
1 È
1
1
dx - Ú
dx ˙ = Kt
ÍÚ
(a - b) ÍỴ 0 (b - x)
(a - x) ˙˚
0

fi

1 È
b
a ˘
ln
- ln
= Kt
(a - b) ÍỴ b - x
a - x ˙˚

fi

1
(a - x) ˘
È b

ln
◊
= Kt
(a - b) ỴÍ (b - x)
a ˚˙

45°
ln(1/K)

Special case
If concentration of A is too greater than B.
[A] >> [B]
a–x a

ln a

ln t1/2 = ln(1/K) + ln(1/a)
ln t1/2 = ln(1/K) – ln a
Characteristics of second order reaction:
A
h
P
(1) Rate = K[A]2
1
(2) t1/2 μ 2 -1
a
A
h
P
t=0

100
–
t = 10
50
50
t = 30
25
75
t = 70
12.5
87.5
(3) At equal time interval, concentration of reactants are in
H.P.
1
1
(2)
K (t ) =
[ A]t [ A]0
1
1
[ A]2t [ A]0

(1)

1
1
K (3t ) =
[ A]3t [ A]0

(3)

K (2t ) =

IIT JEE PC-V2_01.indd 6

fi

1
È (a - x) - (b - x) ˘
dx = K (t - 0)
(a - b) Ú0 ÍỴ (a - x)(b - x) ˙˚

Slope = 1/K

time
1
since a 1– x = Kt +
a

t

dx
Ú (a - x)(b - x) = K Ú dt
0
0

1/[A]t

1
b

K ¢ = ln
t b-x

1st order w.r.t. [B]

Rate = K[A]0[B]1
H+

H2O + CH3COOC2H5 ỉỉỈ EtOH + CH3COOH
(excess)

Rate = K¢[ester]1 (Pseudo unimolecular)

Nth order
t=0
t=t

A
a
a–x

h

P
–
x

dx
= K (a - x) n
dt

x

t

dx
Ú (a - x)n = K Ú dt
0
0
Kt =

1 È
1
1 ˘
- n -1 ˙
Í
n -1
(n - 1) Ỵ (a - x)
a ˚

5/13/2017 10:25:51 AM


1.7

Chemical Kinetics

Half-life of nth order reaction
1 È
1
1 ˘

Kt1/2 =
(n - 1) ÍỴ (a - a /2) n -1 a n -1 ˙˚

(2) Using integrated rate equation (Hit and Trial)
A
Ỉ
P
t=0
a
–
t1
a – x1
x1
t2
a – x2
x2
:
:
:
:
Zero order reaction
x
x
x
x
K= fiK= 1 = 2 = 3
t
t1 t2 t3

C

1
2 n -1 - 1
= n -1
K (n - 1) a n -1
a

t1/2 =

2 n -1
1
for n π 1\ t1/2 μ n -1
K (n - 1)
a

C=

Graph related to nth order reaction
(i)

1
1
= (n - 1) Kt +
n -1
[ A]t
[ A]0n -1
1
[A]tn–1

K=

(ii) t1/2 = C ◊

1È 1
1 ˘ if K = K = K … second order
K= Í
1
2
3
t Ỵ a - x a ˙˚
(3) Graphical method using concentration

Slope = (n – 1)K
1
[A0]n–1

2.303
a
if K1 = K2 = K3 … first order
log
t
a-x

ln[A]

[A]

time

1
a

n -1

I

O

t

t

t1/2

Slope = C =

1/[A]n – 1

1/[A]t

2n–1 – 1
K(n – 1)

nth order

II

1
an–1

t

(iii) ln t1/2 = ln C – (n – 1) ln a

t

(4) Graphical method using half-life

ln t1/2

ln t1/2

Slope = 1 – n
ln C

Slope = 1 – n
ln a

ln a

Prediction of order of the reaction
(1) Initial rate method or different method
A+BhP
dx
Rate =
= K [ A]m [ B]n
dt
Exp. No.
1.
2.
3.

4.

[A]0
x
2x
2x
p

[B]0
y
y
2y
q

z = K(x)m (y)n
m

t1/2 =

Initial Rate (M sec–1)
z
4z
8z
x
(i)

n

4z = K(2x) (y)

(ii)

8z = K(2x)m (2y)n
m = 2, n = 1
Rate = K[A2][B]

IIT JEE PC-V2_01.indd 7

(5) Half-life method
For nth order reaction:
C
a n -1

t1 Ê a2 ˆ
=
t2 ÁË a1 ˜¯

n -1

t1
ln t1 - ln t2
t2
= n -1fi n =1+
a2
ln
a2 - ln a1
ln
a1
ln

Concentration terms replaced by other terms in first
order integrated equation.
Ist order

t=0

A ỉỉỉỉ
ỈP
aM
–

t=t

a–x

(iii)

x

5/13/2017 10:25:54 AM


1.8

Physical Chemistry-II for JEE (Main & Advanced)

[ A]0 (n A )0 (WA )0 (No. of A) at t = 0
a
=
=

=
=
a - x [ A]t (n A )t (WA )t
(No. of A)at t = t
=

( PA )0
for gas since P μ n (If V and T const.)
( PA )t

(1) Concentration term replaced by partial pressure
A(g) Ỉ B(g) + C(g)
t=0
P0
–
–
t=t
P0 – p
p
p
PT = (P0 – p) + p + p
PT = P0 + p
p = PT – P0
\
P° – p = P0 – (PT – P0) = 2P0 – PT
P0 μ a and P0 – P μ (a – x)
K=

P0
2.303

log
t
2 P0 - PT

P0
P0
a
=
=
a - x P0 - p 2 P0 - PT
(2) Concentration term replaced by volume of reagent
used in titration
Stdandard KMnO4
V0 V0 μ a
Vt μ a – x

V0 μ H+
Vt μ H + + x
V μ H+ + a
V – V0 μ a
V – Vt μ a – x
V - V0 ˘
2.303
È a
=
log Í
K=
t
a
x

V - Vt ˙˚
Ỵ

H2O2

(v)

(4) Inversion of cone sugar
optical rot. (+ r10 )
C12H22O11 +
H2O h
t=0
t=t
tỈ

a
a–x
–

–
–
–

+ r20
- r30
C6H12O6 + C6H12O6
(G)
(F)
–
–

x
x
a
a

| r20 | < | r30 |
r 0 = ar10
r = (a t

(i)
x)r10

+

xr20

-

xr30

(ii)

r = ar20 - ar30
From (i) and (ii); (i) and (iii)
-r t t + r 0
= x and
r10 - r20 + r30
a-x=

Vml sample

(i)
(ii)
(iii)
(iv)

(iii)

r0 - r
=a
r10 - r20 + r30

rt - r
+ r30

r10 - r20
0

a
r -r
=
a - x rt - r

1st order

2H 2O 2 ỉỉỉỉ
Ỉ 2H 2O + O 2
t=0
a
–

t=t
a–x
x
V0
2.303
K=
log
t
Vt
V0 Æ Vol. of KMnO4 used in titration against V ml
sample of Solution of H2O2 at t = 0.
(3) Hydrolysis of ester in acidic medium
H+

CH3COOC2H5 + H2O ỉỉỈ CH3COOH + EtOH
a
–
–
–
a–x
–
x
x
–
–
a
a

k=

2.303
r0 - r
log t
t
r -r

Parallel Reaction
A
A

K1
K2

K1

B
OR A
C

Rate of decomposition of ‘A’,
= [K1 + K2][A]

B

K2
C

- d [ A]
= K1[A] + K2[A]
dt

1 [ A]
K1 + K 2 = ln 0
t [ A]t
[ A]t = [ A]o ◊ e -[ K1 + K 2 ] ¥ t

V0

Std. NaOH

V m1

IIT JEE PC-V2_01.indd 8

d [ B]
= K1[ A]
dt
d [C ]
= K 2 [ A]
Rate of formation of ‘C’=
dt
d [ B ] K1
[ B ] K1
=
=
fi
=
d [C ] K 2
[C ] K 2
Rate of formation of ‘B’=

5/13/2017 10:25:56 AM


Chemical Kinetics

% yield of B =

K1
¥ 100
K1 + K 2

% yield of C =

K2
¥ 100
K1 + K 2

- K1[ A]0
K 2 - K1
Concentration of [B] at any time

at t = 0, [B] = 0 fi C =

e K2t [ B ] =

[A]0 = [A]t + [B]t + [C]t

A

2B

or A

K2

A

K1

2B

K2

d [ B] K1[ A]0
=
[- K1e - K1t + K 2e - K 2t ] = 0
dt
K 2 - K1

3C

3C

K1 and K2 are dissociation constants of A
d [ A]
= [ K1 + K 2 ][ A]
dt
[ B]t [C ]t
+

[ A0 ] = [ A]t +
2
3
d [ B]
= 2 K1[ A] = K B [ A]
dt
dC
= 3K 2 [ A]
dt
2 K1
[ B] 2 K1
=
¥ 100
and % B =
[C ] 3K 2
2 K1 + 3K 2

fi

K1e- K1t = K 2 e- K 2t
K
e( K 2 - K1 )t = 2
K1

fi

tmax =

-

%C =

3K 2
¥ 100
2 K1 + 3K 2

K

K
1
ln 2
K 2 - K1 K1

If K2 >> K1, then concentration of [B] is very small and
practically becomes constant.
d [ B]
= 0 fi K1[ A] - K 2 [ B] = 0
dt
K
fi
[ B] = 1 [ A]
K2
Reversible reaction
H 2 + I2

Reaction in series

K1[ A]0 e( K 2 - K1 )t
K [ A]
- 1 0

K 2 - K1
K 2 - K1

K1[ A0 ] - K1t
- e - K2t ]
[e
K 2 - K1

[ B] =

eg:
K1

Kf
Kb

2HI

r = K f [H 2 ][I 2 ] - Kb [HI]2

K

1
2
A ỉỉ
Ỉ B ỉỉỈ
C

- d [ A]
= K1[ A]

Rate of decomposition of A =
dt
Rate of formation of B = K1[A] – K2[B]
Rate of formation of C = K2[B]
[Conc]

C

1 d [HI]
2 dt
d [HI]
= 2[ K f [H 2 ][I 2 ] - Kb [HI]2 ]
dt
d [H 2 ] d [I 2 ] d [HI]
r=
=
=
dt
dt
2dt
Kinetics for reversible reaction
When I order opposed by I order
r=

B

A

A
t

dB
dB
= K1[ A] - K 2 [ B] fi
+ K 2 [ B] = K1[ A]0 e- K1t
dt
dt
On multiplying by e K2t on both sides
d [ B]
+ K 2e K2t [ B] = K1[ A]0 e( K2 - K1 )t
e K2t ◊
dt
d [e K2t [ B]]
= K1[ A]0 e( K2 - K1 )t
dt
d [e K 2t [ B]] = K1[ A]0 e( K 2 - K1 )t ◊ dt
On integration
e K2t [ B] =

IIT JEE PC-V2_01.indd 9

1.9

K1 ◊ [ A]0 ( K 2 - K1 )t
e
+C
K 2 - K1

Kf
Kb

B

t=0

a

–

t=t

a–x

x

at equilibrium

a – xe

xe

d [ A] - d [a - x] dx
=
=
dt
dt
dt
dx
= K f (a - x) - Kb ( x)
dt

dx
=0
at equilibrium
dt
Kf(a – xe) = Kbxe
-

Kb =

K f (a - xe )
xe

(i)

5/13/2017 10:25:57 AM


1.10

Physical Chemistry-II for JEE (Main & Advanced)

Kbxe = Kfa – Kfxe

Activation energy (Ea)
It is the extra amount of energy required by reactant molecules to reach up to the threshold energy.

(Kf + Kb)xe = Kfa
K f + Kb =

Kf ◊a

(ii)

xe

From Eq. (i),

Ea

E

K f (a - xe ) ◊ x
dx
= K f (a - x) dt
xe
dx K f
=
[axe - xxe - ax + xxe ]
dt
xe
dx È K f ˘
a ˙ ( xe - x)
=Í
dt Ỵ xe ˚
dx
= ( K f + Kb )( xe - x)
dt

Ea = Ethreshold – ER

(i) Physical state
Solid < Liquid < gas

–ln(xe – x) = (Kf + Kb) t –ln xe
xe
1
(K f + K b ) = ln
t (xe - x )
Arrhenius collision theory
1. This theory is mainly applied to bimolecular collision.
2. When reactant molecules collide among themselves
then, only they can convert themselves into product if
Effective collision occur.
3. Collision frequency is generally very high but number
of effective collision or active molecules are comparatively low.
4. For any collision to be effective, there are two barriers.
(a) Energy barriers
In order to have effective collisions, reactant molecules must
possess some minimum amount of energy known as threshold energy.
Fraction
of molecules

Ethreshold
Energy Ỉ

Shaded area represents the fraction of active molecule
which are having energy ≥ Ethreshold
f = e–Ea/RT [fraction of active molecules]
R Ỉ Gas constant

Reaction Coordinate Ỉ

Factor affecting rate of reaction

C = –ln xe

Ea Ỉ activation energy

DH

(b) Orientation Barrier
In order to have effective collisions, the reactant molecules
must collide in the proper direction.

–ln(xe – x) = (Kf + kb)t + C
t = 0;

EP
ER

= Eactivated complex – ER

On integration,
dx
Ú ( xe - x) = ( K f + Kb )Ú dt

at

Activated complex

Ethres.

(ii) Particle size
Smaller is the size if particle, more will be the rate of reaction
because effective surface area increases.
(iii) Temperature
For both endo- and exo-thermic reactions, rate of reaction
increased on increasing the temperature.
On increasing the temperature by 10°C generally rate of
reaction becomes 2 to 3 times.
Rate = z ¥ e–Ea/RT
zμ T
On increasing the temperature by 10°C, rate of reaction
changes mainly because of fraction of effective collisions,
i.e., e–Ea/RT which becomes almost 2 to 3 times.
Temperature coefficient (m)
It is the ratio of two rates when temperature is increased by
10°C.
R
K
m = t +10∞C = t +10∞C = 2 to 3.
Rt ∞C
K t ∞C
DT
R2 K 2
=
= ( m ) 10
R1 K1

Effect of temperature on rate constant (K)

A/C to Arrhenius equation;
K = Ae–Ea/RT
A Ỉ Arrhenius constant or pre-exponential factor or frequency factor.
Ea
ln K = ln A RT
Ea
log K = log A 2.303 RT

T Ỉ temperature

IIT JEE PC-V2_01.indd 10

5/13/2017 10:25:59 AM


Chemical Kinetics

1.11

For example manufacture of NH3 by Haber’s process:
Energy

Ea

Ea¢

+ve catalyst
Ea > EaÂ

Fe(S)

ặ 2NH3(g)
N2(g) + 3H2(g) ổổổ
(3) Auto Catalysts:
In these reactions, one of the products formed act as a
catalyst during the reaction.
H + (aq)

Reaction Coordinate Ỉ

log K 2 = log A -

Ea
2.303 RT2

log K1 = log A -

Ea
2.303 RT1

log

K2
Ea È 1
1˘
=
- ˙
Í
K1 2.303R Ỵ T1 T2 ˚

Note:
1. As the activation energy of reaction increases, rate of
reaction decreases.
2. For two different reactions:
Ea1 > Ea2
DT fi identical (T1 = T2)
fi

Ê K2 ˆ Ê K2 ˆ
ÁË K ˜¯ > ÁË K ˜¯
1 1
1 2

Catalyst
Positive catalyst speed up the reaction by providing alternating path of both for the reaction. In which energy of activated
complex is lesser, so activation energy is lesser. Hence, rate
of reaction increases.

Ỉ
CH3COOC2H5(aq.) + H2O ææææ
CH3COOH(aq.) + C2H5OH(aq.)
(autocatalyst)
Photochemical reactions
These are reactions which occur in presence of light and radiation.
hv
For example H 2 + Cl2 ỉỉỈ 2HCl
These reactions follow zero order kinetics.
rμI
I = intensity of light
r=f◊I

f = quantum efficiency or quantum yield.
Number of moles of reactant reacted
f=
Number of moles of photon absorbed
Reaction mechanism
In order to find the rate expression from the given reaction
mechanism, there are two methods:
(1) R.D.S. method (Rate determining step method)
(a) Select the R.D.S. from reaction mechanism which
is slowest step.
(b) Write the rate expression from the R.D.S. taking it
as elementary.
(c) If there is any intermediate then, remove it.
For example,
Ex (1) Reaction: 2NO2 + F2 h 2NO2F
Mechanism:
K

Energy

Ea

EaÂ

+ve catalyst
Ea > EaÂ

Reaction Coordinate ặ

1

ặ NO2F + F; (slow)
NO2 + F2 ỉỉ

K

2
F + NO2 ỉỉỈ
NO2F; (fast)
r = K1[NO2][F2]
(2) Reaction: 2O3 h 3O2
Mechanism:

Negative catalyst or inhibitor decrease the rate of reaction
by providing an alternating path in which activation energy
increases.

O3

Types of catalysis
(1) Homogenous catalysis:
When reactants and catalysts are present in the same
phase.
For example by lead chamber process

r = K[O3][O]

NO(g)

2SO2(g) + O2(g) ỉỉỉỈ 2SO3(g)
(2) Heterogeneous catalysis:

When reactants and catalyst are present in different phase.
For example Ỉ manufacture of H2SO4 by contact process
V O (s)

2 5
Ỉ 2SO3(g)
2SO2(g) + O2(g) ỉỉỉỉ

IIT JEE PC-V2_01.indd 11

KC

O2 + O;
K

O3 + O ỉỉỈ 2O 2 ;

(fast)
(slow)
(i)

[O 2 ][O]
[O3 ]
here (O) is intermediate.
K [O ]
[O] = C 3
\
[O 2 ]
From (i)
KC =

r = K ◊ KC

[O3 ]2
[O 2 ]

r = K ◊ KC [O3 ]2 [O 2 ]-1

5/13/2017 10:26:00 AM


1.12

Physical Chemistry-II for JEE (Main & Advanced)

(2) Steady state approximation method
This method is based on the fact that net rate of formation of intermediate is zero.
1 - d [NO 2 ]
¥
dt
2
- d [NO 2 ]
= K1[NO 2 ][F2 ] + K 2 [NO 2 ][F]
dt
According to steady state approximation method.
d [F]
= 0 = K1 ◊ [NO 2 ] ◊ [F2 ] - K 2 [NO 2 ] ◊ [F]
dt
fi K1 ◊ [NO2] ◊ [F2] = K2[NO2] ◊ [F]
r=

d [NO 2 ]
= 2 K1[NO 2 ] ◊ [F2 ]
dt
1 d [NO 2 ]
fi
= K1[NO 2 ] ◊ [F2 ]
r= ¥
2
dt
Reation of fractional order:
(i) H2 + Br2 h 2HBr Rate = K [H2] [Br2]1/2
(ii) COCl2 h CO + Cl2 Rate = K [COCl2]3/2
(iii) CO + Cl2 h COCl2 Rate = K [CO]2 [Cl2]1/2
fi

-

(iv) Para H2 h Ortho H2 Rate = K [PH2 ]
(v) Thermal decomposition of acetaldehyde
Rate = K [CH3CHO]3/2

Velocity: a < b < g
Penetration power:
Ionisation power:

Radioactive disintegration
All radioactive decay follow first order reaction:
Rate of disintegration or activity (A)
- dN

=
μN
dt
Where N = (Number of radioactive nuclei left at given time)
- dN
= lN
dt
N = N0e–lt
N = Number of radioactive nuclei taken initially
N

t

1 N
l = ln 0
t N

3/2

RADIOACTIVITY

1 A
l = ln 0
t
A
n

1 m
Ê 1ˆ
l = ln 0 and Nt = N o Á ˜

Ë 2¯
t m
m0 : mass of nuclei initially n = Number of half-lite used
\

Nt
m È1˘
=
=
N 0 m0 ÍỴ 2 ˙˚

n

It is a spontaneous nuclear phenomena in which certain radiations like a, b, g are emitted by the nuclei of radioactive
substances.
Radioactivity is independent of physical conditions like
temperature, pressure etc.

Half-life period (t1/2):
0.693
t1/2 =
l
ttotal = n ¥ t1/2

a-rays
These consist of helium nuclei (He+2). Due to each a-particle
decay, atomic mass number decreases by 4 whereas atomic
number decreases by 2.

Average Life: (tavg.)

t
1
tavg. = = 1/2
l 0.693
= 1.44 Ơ t1/2

A
Z X

ổổ
ặ

A- 4
Z -2

Y +a

b-rays
These are composed of electrons.
0
–1e fi A = 0, Z = –1
1
Ỉ 1P1 + -1e0
0 n ææ

During the b-decay, there is no change in mass number but
atomic number increases by 1 due to the conversion of neutron into proton and electron.
g-rays
These are electro-magnetic radiation. Due to their emission,
there is no change in mass number and atomic number decreases the energy level in nucleus.

IIT JEE PC-V2_01.indd 12

aa>b>g

N = N 0 e - lt
t = tavg =

1
l

N0
= 0.37 N 0
e
It is time in which 37% of the initial radioactive nuclei
remain, i.e., 63% decay.

at

N=

Units of activity
SI units: Disintegration per second (dps) or Becquerel (Bq)
Other units:
Curie (Ci) fi 1 Ci = 3.7 ¥ 1010 dps
Rutherford (Rd) fi 1 Rd = 106 dps
Specific activity
It is the activity of 1 g radioactive substance.

5/13/2017 10:26:01 AM


Chemical Kinetics

Definitions
(1) Isotopes : Same atomic number but different mass number
(2) Isobars : Same mass number but different atomic number
(3) Isotopes : Same number of neutrons : [A – Z] Æ constant
(4) Isodiaphers : (n – p) fi same OR
(A – 2Z) fi same
Radioactive disintegration series
There are four natural disintegration series.
Series
Name
Starting
Element
Thorium series
Th – 232
4n
4n + 1 Neptunium series Np – 237
U – 238
4n + 2 Uranium series
U – 235
4n + 3 Actinium series

Stable End
Product
Pb – 208
Bi – 209

Pb – 206
Pb – 207

Application of radioactivity
Age determination of minerals, rocks or the earth for age
determination, minerals or rock sample is analysed for the
amount of radioactive substance and its stable end product.
When age of the mineral can be obtained from the formula.
1
a
l = ln
t a-x
t = age of mineral
a = initial amount of radioactive substance
Age determination of dead animals and plants or carbon
dating
14
1
14
1
7N + 0n h 6C + 1P
This method is based on ratioactive decay of 14C, which formed
in the upper part of atmosphere according to above reaction.
At the same time, C14 disintegrates so that ratio of radioactive
carbon and non-radioactive carbon becomes constant. This
ratio is available in all living animals and plants. After the
death of animal or plant, this ratio changes due to disintegration of C14. The age can be determined from the formula:
1
a
l = ln

t a-x
Accuracy for this method is not good, when it is applied for
very small or long time period.
Cause of radioactivity
Those nuclei whose n/p ratio is in the range 1 to 1.5 are generally stable. This range is called stability belt.
For unstable nuclei, disintegrate in order to reach to the
stability best.
n

20

IIT JEE PC-V2_01.indd 13

p

1.13

(1) Up to Z = 20; n/p ratio is 1 for most stable nuclei but
above Z = 20, n/p ratio for stable nuclei increases because
as the number of proton increases, electrostatic repulsion between these increases. In order to overcome these repulsive
forces, neutrons increase number.
Above Z = 83; there is no stable nuclei.
Among the stable nuclei, n/p ratio is maximum for Bi bismuith (1.5).
Those nuclei which have number of neutron protons equal
to magic number Ỉ 2, 8, 20, 28, 50, 82, 126 are more stable.
Maximum number of stable nuclei have even number of
neutrons and even number of protons whereas very few stable nuclei have odd noumber of neutron and proton.
Favourable condition for disintegration
Type of
decay

Favourable
condition

1. a-decay
for
2. b-decay

Effect

Z > 83

n
≠
p

(heavy unstable nuclei

92 U

high n/p

n
Ø
p
14

6C

238

ỉỉ
Ỉ 90Th 234 + 24a

h 7N14 + –1e0

n
n
= 1.33, = 1.0
p
p
3. g-decay

4.

high-energy decrease the energy level.
level in nuclei

Positron
decay

low

n
ratio
p

n
≠
p

h 6C13 + 1e0
n/p = 6/7
n/p = 7/6
5. k-capture 1P1 + –1e0 h 0n1
(b+ decay)

7N

13

It is the phenomena in which lighter unstable nuclei having
low n/p ratio capture the e– from the nearest shell (i.e., Kn
shell) in order to increase the
ratio.
p
7
4 Be

+ -1e0 ỉỉ
Ỉ 37 Li

n 3
=
p 4

n 4
=
p 3

Nuclear reactions
Reactant nuclei + Bombarding particle Ỉ product nuclei
+ emitted particle
Representation
Reactant nucleus (bombarding particle, emitted particle)
product nucleus.
24
1
24
1
11Na + 1P h 12Mg + 0n
24

11Na

(1P1, 0n1) 12Mg24

5/13/2017 10:26:03 AM


1.14

Physical Chemistry-II for JEE (Main & Advanced)

Example

E = Dmc2

7
4 3

8
3 Li (a 2 , 2 He)3 Li

Example 94 Be(g 0 n1 )84 Be
During the nuclear fission and fusion, large amount of energy
is released because of mass defect.
Dm = mass defect = mR – mP

1 amu = 931.5 MeV
A1 - A2
4
No of b particle emitted = 2a – [Z1 – Z2]
Number of a particle emitterd =

Solved Examples
1. At 10°C for any reaction rate constant = K, what will be
rate constant at 90°C (given temperature coefficient 2)
Sol.

[NH3] 0.6

DT
R2 K 2
=
= ( m ) 10 = 28 K
R1 K1

0.4
0.2

2. (a) For the following reaction write down rate law:
N2 + 3H2 h 2NH3
(b) If rate of dissappearance H2 is
the rate of appearance of NH3.

12 g
. What is
- sec

1 Ê - d [N 2 ] ˆ 1 Ê - d [H 2 ] ˆ
Sol. (a) Rate = Á
˜= Á
˜
1 Ë dt ¯ 3 Ë dt ¯
1 + [dNH3 ]
=
= K [N 2 ]m [H 2 ]n
2
dt
(b)

2 Ê - d [H 2 ] ˆ 2 12
moles
=4
ÁË
˜¯ = ¥
dt
- sec
3
3 2

gram
= 4 ¥ 17 = 68 g
- sec

d [NH3 ]
=
dt

3. For the reaction, 2A Ỉ 3B + 4C, the number of moles
of B increases by 6 ¥ 10–3 moles in 10 sec in a 10 L
container. Calculate:
(a) rate of appearance of B and C.
(b) rate of disappearance of A
Sol. (a) rB =

DConc. 6 ¥ 10-3
=
= 6 ¥ 10-5
Dt
10 ¥ 10

rC rB
4
= fi rC = ¥ rB = 8 ¥ 10-3
4
3
3
(b)

rA rB

2
= fi rA = rB = 4 ¥ 10-5
2
3
3

4. For the reaction; 2NH3 Ỉ N2 + 3H2, the curve is plotting between concentration of NH3 v/s time. Calculate
(a) rate of decomposition of NH3 between 5 to 10
seconds.
(b) rate of reaction between 10 to 20 seconds.

5

10 20
Time (sec) Æ

D[NH3 ] 0.2
=
= 0.04 M/sec -1
Dt
5
1 0.2
r= ¥
= 0.01 M/sec -1
2 10
5. 2N2O5 h 4NO2 + O2, the variation of concentration of N2O5 with time can be expressed by [N2O5] =
[N2O5]0 e–Kt and K = 10–4 sec–1. If initially 1 mole of
N2O5 taken calculate.
(i) Rate of disappearance of N2O5 at t = 104 sec.
(ii) Rate of reaction during first 104 sec.

(iii) Variation of concentration of NO2 with time.
-

(i) Rate = -

d [N 2O5 ]
= K [N 2O5 ]0 e - Kt
dt

= K [N 2O5 ]0 e -10

-4

¥104

1
e
1
= 1 ¥ 10-4 [1] ¥
e
= K [N 2O5 ]0 ¥

x
= 1 - e - Kt for first order.
a
1 - 1/e 1
(ii) R =
¥
2
104

4NO2
+
O2
(iii) 2N2O5 Ỉ
1
–
–
1–a
2a
a/2
We know a =

2 mole

[NO2]

Time

IIT JEE PC-V2_01.indd 14

5/13/2017 10:26:04 AM