NCERT Solutions for Class 12 Chemistry Chapter 4 Chemical kinetics
NCERT solutions for class 12 chemistry chapter 4 Chemical Kinetics  Chemical kinetics deals with the average and instantaneous rate of reaction, factors affecting these and the mechanism of the reaction. Chemical kinetics helps students to understand how chemical reactions occur. In NCERT solutions for class 12 chemistry chapter 4 Chemical Kinetics, there are questions and solutions of some important topics like average and instantaneous rate of a reaction, fact ors affecting the rate of reaction, the integrated rate equations for the zero and firstorder reactions, etc. This chapter also tells you when is a chemical reaction feasible and how can we calculate the speed of reactions.
This chapter holds 5 marks in the class 12 CBSE board exam of chemistry. In this chapter, there are 30 questions in the exercise and 9 questions which are related to topics studied. To clear doubts of students, the CBSE NCERT solutions for class 12 chemistry chapter 4 Chemical Kinetics are prepared in a comprehensive manner by subject experts. This chapter is vital for both CBSE Board exam as well as for competitive exams like JEE Mains, VITEEE, BITSAT, etc. so, students must pay special attention to concepts of this chapter. The NCERT solutions provided here are completely free and you can also download them for offline use also if you want to prepare or any other subject or any other class.
Rate of reaction It is defined as the rate of change in concentration of reactant or product. Unit of rate is .
Reactants, R Products, P
nA+mB pC+qD
Topics and Subtopics of NCERT Class 12 Chemistry Chapter 4 Chemical Kinetics
4.1Rate of a Chemical Reaction
4.2Factors Influencing Rate of a Reaction
4.3Integrated Rate Equations
4.4Temperature Dependence of the Rate of a Reaction
4.5 Collision Theory of Chemical Reactions
NCERT solutions for class 12 chemistry chapter 4 Chemical Kinetics 
Solutions to In Text Questions Ex 4.1 to 4.9
Answer :
We know that,
The average rate of reaction =
=
=
=
In seconds we need to divide it by 60. So,
=
= 6.67
Question 4.2 In a reaction, P, the concentration of A decreases from to in 10 minutes. Calculate the rate during this interval?
Answer :
According to the formula of an average rate
= (final concentration  initial conc.)/time interval
=
=
=
=
Question 4.3 For a reaction, ; the rate law is given by, . What is the order of the reaction?
Answer :
Order of reaction = Sum of power of concentration of the reactant in the rate law expressions
So, here the power of A = 0.5
and power of B = 2
order of reaction = 2+0.5 =2.5
Answer :
The order of a reaction means the sum of the power of concentration of the reactant in rate law expression.
So rate law expression for the secondorder reaction is here R = rate
if the concentration is increased to 3 times means
new rate law expression = = = 9R
the rate of formation of Y becomes 9 times faster than before
Question 4.5 A first order reaction has a rate constant . How long will of his reactant take to reduce to ?
Answer :
Given data,
initial conc. = 5g
final conc. = 3g
rate const. for firstorder =
We know that for the firstorder reaction,
[log(5/3)= 0.2219]
= 444.38 sec (approx)
Answer :
We know that t(half ) for the firstorder reaction is
and we have given the value of half time
thus,
= 0.01155 /min
OR = 1.1925
Alternative method
we can also solve this problem by using the firstorder reaction equation.
put
Question 4.7 What will be the effect of temperature on rate constant ?
Answer :
The rate constant of the reaction is nearly doubled on rising in 10degree temperature.
Arrhenius equation depicts the relation between temperature and rates constant.
A= Arrhenius factor
Ea = Activation energy
R = gas constant
T = temperature
Question 4.8 The rate of the chemical reaction doubles for an increase of in absolute temperature from Calculate .
Answer :
Given data
(initial temperature) = 298K and (final temperature)= 308K
And we know that rate of reaction is nearly doubled when temperature rise 10degree
So, and R = 8.314 J/mol/K
now,
On putting the value of given data we get,
Activation energy ( ) =
=52.9 KJ/mol(approx)
Question 4.9 The activation energy for the reaction is at . Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?
Answer :
We have
Activation energy = 209.5KJ/mol
temperature= 581K
R = 8.314J/mol/K
Now, the fraction of molecules of reactants having energy equal to or greater than activation energy is given as
taking log both sides we get
= 18.832
x = antilog(18.832)
= 1.471
NCERT Solutions for class 12 chemistry chapter 4
Question 4.1(i) From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.
Answer :
Given pieces of information
Rate =
so the order of the reaction is 2
The dimension of k =
Question 4.1(ii) From the rate expression for the following reactions determine their order of reaction and the dimensions of the rate constants.
Answer :
Given rate =
therefore the order of the reaction is 2
Dimension of k =
Question 4.1(iii) From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.
Answer :
Given
therefore the order of the reaction is 3/2
and the dimension of k
Question 4.1(iv) From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.
Answer :
so the order of the reaction is 1
and the dimension of k =
Question
4.2
For the reaction:
the rate =
with
. Calculate the initial rate of the reaction when
. Calculate the rate of reaction after
is reduced to
.
Answer :
The initial rate of reaction =
substitute the given values of [A], [B] and k,
rate =
=8
When [A] is reduced from 0.1 mol/L to 0.06 mol/L
So, conc. of A reacted = 0.10.06 = 0.04 mol/L
and conc. of B reacted = 1/2(0.04) = 0.02mol/L
conc. of B left = (0.20.02) = 0.18 mol/L
Now, the rate of the reaction is (R) =
=
Question 4.3 The decomposition of on platinum surface is zero order reaction. What are the rates of production of and if ?
Answer :
The decomposition of on the platinum surface reaction
therefore,
Rate =
For zero order reaction rate = k
therefore,
So
and the rate of production of dihydrogen = 3 (2.5 )
= 7.5
Answer :
Given that
So, the unit of rate is bar/min .( )
And thus the unit of k = unit of rate
Question 4.5 Mention the factors that affect the rate of a chemical reaction.
Answer :
The following factors that affect the rate of reaction

the concentration of reactants

temperature, and

presence of catalyst
Question 4.6(i) A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is doubled
Answer :
Let assume the concentration of reactant be x
So, rate of reaction,
Now, if the concentration of reactant is doubled then . So the rate of reaction would be
Hence we can say that the rate of reaction increased by 4 times.
Question 4.6(ii) A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is reduced to half ?
Answer :
Let assume the concentration of reactant be x
So, rate of reaction, R =
Now, if the concentration of reactant is doubled then . So the rate of reaction would be
Hence we can say that the rate of reaction reduced to 1/4 times.
Answer :
The rate constant is nearly double when there is a 10degree rise in temperature in a chemical reaction.
effect of temperature on rate constant be represented quantitatively by Arrhenius equation,
where k is rate constant
A is Arrhenius factor
R is gas constant
T is temperature and
is the activation energy
Question 4.8 In pseudo first order hydrolysis of ester in water, the following results were obtained:
(i) Calculate the average rate of reaction between the time interval 30 to 60 seconds.
Answer :
The average rate of reaction between the time 30 s to 60 s is expressed as
Question 4.9(i) A reaction is first order in A and second order in B.
(i)Write the differential rate equation.
Answer :
the reaction is first order in A and second order in B. it means the power of A is one and power of B is 2
The differential rate equation will be
Question 4.9(ii) A reaction is first order in A and second order in B.
(ii) How is the rate affected on increasing the concentration of B three times?
Answer :
If the concentration of [B] is increased by 3 times, then
Therefore, the rate of reaction will increase 9 times.
Question 4.9(iii) A reaction is first order in A and second order in B.
(iii) How is the rate affected when the concentrations of both A and B are doubled?
Answer :
If the concentration of [A] and[B] is increased by 2 times, then
Therefore, the rate of reaction will increase 8 times.
What is the order of the reaction with respect to A and B?
Answer :
we know that
rate law (
) =
As per data
these are the equation 1, 2 and 3 respectively
Now, divide eq.1 by equation2, we get
from here we calculate that
y = 0
Again, divide eq. 2 by Eq. 3, we get
Since y =0 also substitute the value of y
So,
=
=
taking log both side we get,
= 1.496
= approx 1.5
Hence the order of reaction w.r.t A is 1.5 and w.r.t B is 0(zero)
Question
4.11
The following results have been obtained during the kinetic studies of the reaction:
2A + B
C + D
Determine the rate law and the rate constant for the reaction .
Answer :
Let assume the rate of reaction wrt A is
and wrt B is
. So, the rate of reaction is expressed as
Rate =
According to given data,
these are the equation 1, 2, 3 and 4 respectively
Now, divide the equation(iv) by (i) we get,
from here we calculate that
Again, divide equation (iii) by (ii)
from here we can calculate the
value of y is 2
Thus, the rate law is now,
So,
Hence the rate constant of the reaction is
Answer :
The given reaction is first order wrt A and zero order in wrt B. So, the rate of reaction can be expressed as;
Rate = k[A]
from exp 1,
k = 0.2 per min.
from experiment 2nd,
[A]
=
from experiment 3rd,
from the experiment 4th,
from here
[A] = 0.1 mol/L
Question
4.13 (1)
Calculate the halflife of a first order reaction from their rate constants given below:
Answer :
We know that,
halflife (
) for firstorder reaction =
=
Question
4.13 (2)
Calculate the halflife of a first order reaction from their rate constants given below:
Answer :
the halflife for the firstorder reaction is expressed as ;
= 0.693/2
= 0.35 min (approx)
Question
4.13 (3)
Calculate the halflife of a first order reaction from their rate constants given below:
Answer :
The halflife for the firstorder reaction is
= 0.693/4
= 0.173 year (approximately)
Answer :
Given ,
halflife of radioactive decay = 5730 years
So,
per year
we know that, for firstorder reaction,
= 1845 years (approximately)
Thus, the age of the sample is 1845 years
Question 4.15 (1) The experimental data for decomposition of
in gas phase at 318K are given below:
Plot against t.
Answer :
On increasing time, the concentration of gradually decreasing exponentially.
Question 4.15 (2) The experimental data for decomposition of in gas phase at 318K are given below:
Find the halflife period for the reaction.
Answer :
The halflife of the reaction is
The time corresponding to the
mol/ L = 81.5 mol /L is the halflife of the reaction. From the graph, the answer should be in the range of 1400 s to 1500 s.
Question 4.15 (3) The experimental data for decomposition of in gas phase at 318K are given below:
Answer :



0 
1.63 
1.79 
400 
1.36 
1.87 
800 
1.14 
1.94 
1200 
0.93 
2.03 
1600 
0.78 
2.11 
2000 
0.64 
2.19 
2400 
0.53 
2.28 
2800 
0.43 
2.37 
3200 
0.35 
2.46 
Question 4.15 (4) The experimental data for decomposition of in gas phase at 318K are given below:
Answer :
Here, the reaction is in first order reaction because its log graph is linear.
Thus rate law can be expessed as
Question 4.15 (5) The experimental data for decomposition of in gas phase at 318K are given below:
Answer :
From the log graph,
the slope of the graph is =
= k/2.303 ..(from log equation)
On comparing both the equation we get,
Question 4.15 (6) The experimental data for decomposition of in gas phase at 318K are given below:
Calculate the halflife period from k and compare it with (ii).
Answer :
The half life produce =
=
Question 4.15(7) The rate constant for a first order reaction is . How much time will it take to reduce the initial concentration of the reactant to its 1/16th value?
Answer :
We know that,
for first order reaction,
(nearly)
Hence the time required is
Answer :
Given,
half life = 21.8 years
= 0.693/21.8
and,
by putting the value we get,
taking antilog on both sides,
[R] = antilog(0.1071)
= 0.781
Thus 0.781 of will remain after given 10 years of time.
Again,
Thus 0.2278
of
will remain after 60 years.
Answer :
case 1
for 99% complition,
CASE II
for 90% complition,
Hence proved.
Question 4.19 A first order reaction takes 40 min for 30% decomposition. Calculate
Answer :
For the firstorder reaction,
(30% already decomposed and remaining is 70%)
therefore half life = 0.693/k
=
= 77.7 (approx)
Question 4.20 For the decomposition of azoisopropane to hexane and nitrogen at 543K, the following data are obtained.
Answer :
Decompostion is represented by equation
After t time, the total pressure
=
So,
thus,
for first order reaction,
now putting the values of pressures,
when t =360sec
when t = 270sec
So,
Question
4.21
The following data were obtained during the first order thermal decomposition of
at a constant volume.
Calculate the rate of the reaction when total pressure is 0.65 atm.
Answer :
The thermal decomposition of
is shown here;
After t time, the total pressure =
So,
thus,
for first order reaction,
now putting the values of pressures,
when t = 100s
when
= 0.65  0.5
= 0.15 atm
So,
= 0.5  0.15
= 0.35 atm
Thus, rate of reaction, when the total pressure is 0.65 atm
rate = k(
)
=
= 7.8
Question
4.22
The rate constant for the decomposition of N2O5 at various temperatures
is given below:
Answer :
From the above data,
T/ 
0 
20 
40 
60 
80 
T/K 
273 
293 
313 
333 
353 
( ) 
3.66 
3.41 
3.19 
3.0 
2.83 

0.0787 
1.70 
25.7 
178 
2140 

7.147 
4.075 
1.359 
0.577 
3.063 
Slope of line =
According to Arrhenius equations,
Slope =
12.30
8.314
= 102.27
Again,