# 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 first-order 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 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.By referring to the NCERT solutions for class 12 , students can understand all the important concepts and practice questions well enough before their examination.

** 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 Sub-topics 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 second-order 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 first-order =

We know that for the first-order reaction,

[log(5/3)= 0.2219]

= 444.38 sec (approx)

** Answer ** :

We know that t(half ) for the first-order 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 first-order 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 10-degree 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 10-degree

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 = 0.1-0.06 = 0.04 mol/L

and conc. of B reacted = 1/2(0.04) = 0.02mol/L

conc. of B left = (0.2-0.02) = 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 10-degree 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 half-life of a first order reaction from their rate constants given below:

** Answer ** :

We know that,

half-life ( ) for first-order reaction =

=

** Question ** ** 4.13 (2) ** Calculate the half-life of a first order reaction from their rate constants given below:

** Answer ** :

the half-life for the first-order reaction is expressed as ;

= 0.693/2

= 0.35 min (approx)

** Question ** ** 4.13 (3) ** Calculate the half-life of a first order reaction from their rate constants given below:

** Answer ** :

The half-life for the first-order reaction is

= 0.693/4

= 0.173 year (approximately)

** Answer ** :

Given ,

half-life of radioactive decay = 5730 years

So,

per year

we know that, for first-order 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 half-life period for the reaction.

** Answer ** :

The half-life of the reaction is-

The time corresponding to the mol/ L = 81.5 mol /L is the half-life 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 half-life 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 first-order 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 ** :

Decomposition 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, ** When T = 30 +273 = 303 K ** and 1/T =0.0033K

** k = **

** When T = 50 + 273 = 323 K ** and 1/T = 3.1 K

** k = 0.607 per sec **

** Answer ** :

Given that,

k =

= 179.9 KJ/mol

T(temp) = 546K

According to Arrhenius equation,

taking log on both sides,

= (0.3835 - 5) + 17.2082

= 12.5917

Thus A = antilog (12.5917)

A = 3.9 per sec (approx)

** Answer ** :

Given that,

k =

t = 100 s

Here the unit of k is in per sec, it means it is a first-order reaction.

therefore,

Hence the concentration of rest test sample is 0.135 mol/L

** Answer ** :

For first order reaction,

given that half life = 3 hrs ( )

Therefore k = 0.693/half-life

= 0.231 per hour

Now,

= antilog (0.8024)

= 6.3445

(approx)

Therefore fraction of sample of sucrose remains after 8 hrs is 0.157

** Question ** ** 4.26 ** The decomposition of hydrocarbon follows the equation . Calculate

** Answer ** :

The Arrhenius equation is given by

.................................(i)

given equation,

............................(ii)

by comparing equation (i) & (ii) we get,

A= 4.51011 per sec

Activation energy = 28000 (R = 8.314)

= 232.792 KJ/mol

** Question ** ** 4.27 ** The rate constant for the first order decomposition of is given by the following equation:

.

Calculate for this reaction and at what temperature will its half-period be 256 minutes?

** Answer ** :

The Arrhenius equation is given by

taking log on both sides,

....................(i)

given equation,

.....................(ii)

On comparing both equation we get,

activation energy

half life ( ) = 256 min

k = 0.693/256

With the help of equation (ii),

T =

= 669 (approx)

** Question ** ** 4.28 ** The decomposition of A into product has value of k as at 10°Cand energy of activation 60 kJ mol–1. At what temperature would k be ?

** Answer ** :

The decomposition of A into a product has a value of k as at 10°C and energy of activation 60 kJ mol–1.

K1 =

K2 =

= 60 kJ mol–1

K2 =

** Answer ** :

We know that,

for a first order reaction-

Case 1

At temp. = 298 K

= 0.1054/k

Case 2

At temp = 308 K

= 2.2877/k'

As per the question

K'/K = 2.7296

From Arrhenius equation,

= 76640.096 J /mol

=76.64 KJ/mol

k at 318 K

we have , T =318K

A=

Now

After putting the calue of given variable, we get

on takingantilog we get,

k = antilog(-1.9855)

= 1.034

** Answer ** :

From the Arrhenius equation,

...................................(i)

it is given that

T1= 293 K

T2 = 313 K

Putting all these values in equation (i) we get,

** Activation Energy = 52.86 KJ/mo ** l

This is the required activation energy

** NCERT Solutions for Class 12 Chemistry **

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** Benefits of NCERT solutions for class 12 chemistry chapter 4 Chemical Kinetics **

- First, the easy steps given in the NCERT solutions for class 12 chemistry chapter 4 will help you to understand the chapter easily.
- Revision will be so easy such that you always remember the concepts and get very good marks in your class.
- Homework will be easy now, all you need to do is check the detailed CBSE NCERT solutions for class 12 chemistry chapter 4 Chemical Kinetics and you are good to go.

If you have a doubt or question that is not available here or in any of the chapters, contact us. You will get the answers with solutions that will help you score well in your exams.

## Frequently Asked Question (FAQs) - NCERT Solutions for Class 12 Chemistry Chapter 4 Chemical kinetics

**Question: **Where can I find complete solutions of NCERT class 12 Chemistry?

**Answer: **

For complete solutions : https://school.careers360.com/ncert/ncert-solutions-class-12-chemistry

**Question: **What are the important topics of this chapter?

**Answer: **

Important topics of this chemical kinetics are

- The rate of reaction
- concept of collision theory
- effect of temperature in activation energy
- Arrhenious equation
- concept of collision theory

**Question: **What is the weightage of NCERT class 12 Chemistry chapter 4 in CBSE board exam ?

**Answer: **

The weightage of NCERT class 12 Chemistry chapter 4 in CBSE board exam is 5 marks

**Question: **What is the weightage of NCERT class 12 Chemistry chapter 4 in NEET?

**Answer: **

The weightage of NCERT class 12 Chemistry chapter 4 in NEET is 3%

**Question: **What is the weightage of NCERT class 12 Chemistry chapter 4 in JEE Mains?

**Answer: **

Weightage of NCERT class 12 Chemistry chapter 4 in JEE Mains is 4 marks

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