narayana

84 Pages·2007·1.01 MB·English

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C H E M I S T R Y
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S T U D Y M A T E R I A L
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PHASE - I
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(cid:4)CHEMICAL KINETICS &(cid:5)
(cid:4) NUCLEAR CHEMISTRY (cid:5)
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(cid:4) IIT-JEE (cid:5)
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NARAYANA INSTITUTE OF CORRESPONDENCE COURSES
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FNS HOUSE, 63 KALU SARAI MARKET
SARVAPRIYA VIHAR, NEW DELHI-110016
PH.: (011) 32001131/32/50 • FAX : (011) 41828320
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E-mail : i n f o @ n a r a y a n a i c c . c o m
 2004 NARAYANA GROUP
This study material is a part of NARAYANA INSTITUTE OF CORRESPONDENCE COURSES for IIT-JEE, 2008-09. This is meant
for the personal use of those students who are enrolled with NARAYANA INSTITUTE OF CORRESPONDENCE COURSES, FNS
House, 63, Kalu Sarai Market, New Delhi-110016, Ph.: 32001131/32/50. All rights to the contents of the Package rest with
NARAYANA INSTITUTE. No other Institute or individual is authorized to reproduce, translate or distribute this material in any form,
without prior information and written permission of the institute.
PREFACE
Dear Student,
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The salient features of this package include :
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Objective - Multiple Choice Questions, Comprehension Type Questions, Match the Following, Assertion-Reasoning
& Subjective Questions.
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THE NARAYANA TEAM
CCCCCOOOOONNNNNTTTTTEEEEENNNNNTTTTTSSSSS
CHEMICAL KINETICS
SSSSS
& NUCLEAR CHEMISTRY
TTTTT 1. Theory
2. Solved Problems
(i) Subjective Type Problems
NNNNN
(ii) Single Choice Problems
(iii) Multiple Choice Problems
(iv) Miscellaneous Problems
EEEEE
• Comprehension Type Problems
• Matching Type Problems
TTTTT • Assertion-Reason Type Problems
3. Assignments
(i) Subjective Questions
NNNNN
(ii) Single Choice Questions
(iii) Multiple Choice Questions
(iv) Miscellaneous Questions
OOOOO
• Comprehension Type Questions
• Matching Type Questions
• Assertion-Reason Type Questions
CCCCC
(v) Problems Asked in IIT-JEE
4. Answers
CHEMICAL KINETICS
& NUCLEAR CHEMISTRY

IIT-JEE Syllabus
Rates of chemical reactions; Order of reactions; Rate constant; First order
reactions; Temperature dependence of rate constant (Arrhenius equation).

CONTENTS INTRODUCTION
(cid:190) Rate of reaction Some chemical reaction occurs within
few micro seconds or milli seconds due
(cid:190) Elementary reaction
to formation of reaction intermediate
(carbocation, carboanion or free radicals).
(cid:190) Integrated rate laws
But some reaction occur in few year or
(cid:190) Activation energy months because breaking of strong ionic or
metallic bond is a slow process. Under
(cid:190) Kinetics of some complex first order
chemical kinetics we would study that
reactions
reaction which occurs at measurable rate.
(cid:190) Methods of determining order
In Chemical Kinetics we would be able to
(cid:190) Nuclear chemistry understand the velocity as well as different
factors which would effect the chemical
(cid:190) Kinetics of radioactive disintegration
reaction. Under this we will be studying
the mechanism of the reaction. Here
mechanism of reaction means how
reactants are converted into product or in
how many intermediate steps reactant is
converted into product. And which
intermediate step is rate determining step.

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NARAYANA
Chemistry : Chemical Kinetics & Nuclear Chemistry
INSTITUTE OF CORRESPONDENCE COURSES

1. RATE OF REACTION
Rate of a reaction is defined as the increase in molar concentration of a product per unit
time or decrease in molar concentration of a reactant per unit time. Molar concentration is
normally measured in moles per litre, the rate of a reaction is specified in mole per litre per
time.
1.1 AVERAGE RATE
∆x
The quantity is known as average rate, where ∆x is the change in concentration in
∆t
∆ttime. As we have discussed in chemical equilibrium rate of change of concentration of
reactant decreases as the reaction proceeds. This means that rate of change of concentration
is not constant. If the time interval is quite large then average rate will show large
deviations from the actual rate.
1.2 INSTANTANEOUS RATE
In order to precisely define rate time interval ∆tis made smaller
i.e. instantaneous rate = average rate as ∆tapproaches zero.
∆[A] d[A]
= =
 
 ∆t  dt
∆t→0
Consider the hypothetical reaction
A+2B→3C+D
d[A]
Rate of disappearance of A = −
dt
d[B]
Rate of disappearance of B = −
dt
d[C]
Rate of appearance of C = +
dt
d[D]
Rate of appearance of D = +
dt
But from the stoichiometry it is apparent that when one mole of A is reacted, two moles of
B are also consumed.
i.e. rate of disappearance of B = 2 × rate of disappearance of A
d[B] d[A]
− =2×−
dt dt
d[A] 1 d[B]
or − =−
dt 2 dt

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NARAYANA
Chemistry : Chemical Kinetics & Nuclear Chemistry

INSTITUTE OF CORRESPONDENCE COURSES
Similarly we can prove that
d[A] 1 d[C]
− =+
dt 3 dt
d[A] 1 d[B] 1d[C] d[D]
So, rate of reaction = − =− = =
dt 2 dt 3 dt dt
We can also generalize our statement i.e. for a general reaction
m A+m B→n C+n D
1 2 1 2
1 d[A] 1 d[B] 1 d[C] 1 d[D]
Rate of reaction = − =− =+ =+
m dt m dt n dt n dt
1 2 1 2
Illustration 1 : Dinitrogen pentaoxide decomposes as follows:
1 d[N O ]
N O →2NO + O . If, − 2 5 = K′[N O ]
2 5 2 2 2 dt 2 5
d[NO ]
2 = K′′[N O ]
dt 2 5
d[O ]
2 = K′′′[N O ]
dt 2 5
Derive a relation in, K′,K'' and K '''
d[N O ] 1 d[NO ] d[O ]
Solution : For the given change − 2 5 =+ 2 =+2 2
dt 2 dt dt
1
On substituting values, K′[N O ]= K′′[N O ]= 2K′′′[N O ]
2 5 2 2 5 2 5
or 2K′=K′′=K′′′
EXERCISE 1
1. For the reaction; 4NH + 5O → 4NO + 6H O , the rate of reaction in terms of
3(g) 2(g) (g) 2 (g)
d[NH ]
disappearance of NH is − 3 , then write rate expression in terms of O , NO and
3 2
dt
H O.
2
2. For the decomposition reaction: N O → 2NO ; the initial pressure of N O falls
2 4(g) 2(g) 2 4
from 0.46 atm to 0.28 atm in 30 minute. What is the rate of appearance of NO ?
2
3. A chemical reaction 2A → 4B + C; in gaseous phase shows an increase in concentration
of B by 5 × 10–3 M in 10 second. Calculate:
a) rate of appearance of B,
b) rate of the reaction,
c) rate of disappearance of A

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NARAYANA
Chemistry : Chemical Kinetics & Nuclear Chemistry
INSTITUTE OF CORRESPONDENCE COURSES

2. ELEMENTARY REACTION
An elementary reaction is a single molecular event, such as collision of molecules resulting
in a reaction. The set of elementary reactions whose overall effect is given by the net
chemical equation is called reaction mechanism.
Let us consider an example for the reaction of nitrogen dioxide with carbon monoxide.
NO +CO →NO +CO [net chemical equation] …(i)
2(g) (g) (g) 2(g)
Suppose this chemical reaction takes place in two steps
NO +NO →NO +NO [Elementary reaction] …(ii)
2(g) 2(g) 3(g) (g)
NO +CO →NO +CO [Elementary reaction] …(iii)
3(g) (g) 2(g) 2(g)
Thus net reaction (i) is obtained by the combination of two elementary reaction (ii) and (iii)
and NO is a reaction intermediate as it is produced and consumed during the course of
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reaction.
2.1 MOLECULARITY
It is defined as the number of molecules taking part on the reactant side of an elementary
reaction. A unimolecular reaction is an elementary reaction involves one reactant molecule,
a bimolecular reaction is an elementary reaction that involves two reactant molecules.
2.2 DIFFERENTIAL RATE LAW
It is a relationship that relates the variation of rate of reaction with the concentration of
reactants.
Consider the reaction
2A+3B→C+D
as we have seen
1 d[A] d[C]
− = ∝[A]m[B]n
2 dt dt
[law of mass action states that rate of reaction is directly proportional to the product of the
active masses of reactants raised to some powers]
d[C]
So, =k[A]m[B]n
dt
k is known as velocity constant or rate constant or specific reaction rate. m is known as
order with respect to A, n is known as order with respect to B. Sum m + n is known as
overall order of the reaction.
2.3 ORDER OF REACTION
It is defined as the sum of exponents or powers which are raised to concentration terms in
the rate law expression of a reaction.

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NARAYANA
Chemistry : Chemical Kinetics & Nuclear Chemistry

INSTITUTE OF CORRESPONDENCE COURSES
Difference between order and Molecularity : Molecularity of a reaction can be used to
describe only an elementary process and molecularity can be predicted just be viewing the
elementary reaction. Whereas order of a reaction refers to the overall reaction and can be
determined experimentally, however for elementary process the order and molecularity
are same. Molecularity has got no meaning for an overall reaction and order cannot be
predicted from a balanced chemical equation.
Note : If a reaction can be written as a combination of several elementary reaction then it’s the
slowest step that governs the rate of reaction i.e. the slowest step is the rate determining step.
Differential Rate Law :
Consider reaction
P+2Q→R
The differential rate law is written as
d[P] 1 d[Q]
Rate=− =− =k[P]m[Q]n
dt 2 dt
Value of m and n can be determined by performing the reaction in laboratory i.e. order
w.r.t P is m and order w.r.t. Q is n and overall order will be m + n.
4.4 UNITS OF k
In general rate law for a nth order reaction can be written as
dC
=kCn
dt
Where k is rate constant, k is characteristic of a reaction at a given temperature. It changes
only when temperature changes and n is the order of reaction.
dC/dt
k =
Cn
Units of k: (concentration)1–ntime–1
For a zero order reaction (n = 0)
Units of k = mol/L/s
For a first order reaction (n = 1)
Units of k = time–1
For a second order reaction (n = 2)
Units of k = (mol/l)–1time–1 = l/mol/s

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FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 (cid:127) Ph.: [phone]/32 Fax : [phone]
NARAYANA
Chemistry : Chemical Kinetics & Nuclear Chemistry
INSTITUTE OF CORRESPONDENCE COURSES

Illustration 2 :The data given below are for the reaction of NO and Cl to form NOCl at 25K
2
[Cl ] [NO] Initial rate × 103 (mol litre–1sec–1)
2
0.05 0.05 1
0.15 0.05 3
0.05 0.15 9
a) What is the order with respect to NO and Cl in the reaction?
2
b) Write the rate expression.
c) Calculate the rate constant
d) Determine the reaction rate when conc. of Cl and NO are 0.2 M and 0.4 M
2
respectively.
Solution : For the reaction; 2NO + Cl → 2NOCl
2
Rate = k[Cl ]m[NO]n …(1)
2
Where, m and n are order of reaction w.r.t Cl and NO, respectively.
2
From the given data:
1 × 10–3 = k[0.05]m [0.05]n …(2)
3 × 10–3 = k [0.15]m [0.05]n …(3)
9 × 10–3 = k[0.05]m [0.15]n …(4)
By equations (2) and (3),
m = 1
By equations (2) and (4)
n = 1
a) ∴ order with respect to NO is 2 and w.r.t. to Cl is 1.
2
b) Also, rate expression r = k[Cl ]1 [NO ]2
2 2
r 1×10−3
c) And rate constant.,k = = = 8 litre2 mol–2 sec–1
[Cl ][NO]2 [0.05]1[0.05]2
2
d) Further, r = k[Cl ]1 [NO]2 = 8 [0.2]1 [0.4]2
2
= 0.256 mol litre–1 sec–1

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