# NCERT solutions for class 12 chemistry chapter 4 Chemical Kinetics

**NCERT Solutions for Class 12 Chemistry Chapter 4 Chemical Kinetics-** Students appearing in the Class 12 board exam must check these NCERT Class 12 Chemistry Chapter 4 solutions. NCERT solutions for Class 12 Chemistry chapter 4 Chemical Kinetics covers all the questions from NCERT books for Class 12 Chemistry.

In NCERT Class 12 Chemistry solutions chapter 4, there are questions and solutions of some important topics like average and instantaneous rate of a reaction, factors affecting the rate of reaction, integrated rate equations for zero and first-order reactions, etc. Read further to know all the Chemical Kinetics Class 12 exercise solutions.

**Also Read,**

- Chemical Kinetics Class 12 NCERT Notes
- Chemical Kinetics Class 12 NCERT Exemplar Solutions

**Check NCERT Class 12 solutions for other subjects.**

** 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

## NCERT solutions for class 12 chemistry chapter 4 Chemical Kinetics -** **

** Question ** ** 4.1 ** For the reaction , the concentration of a reactant change from to in minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.

** 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

** Question ** ** 4.4 ** The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y ?

** 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)

** Question ** ** 4.6 ** Time required to decompose to half of its initial amount is minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.

** 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.

(ii)

** 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

** Question 4.4 **The decomposition of dimethyl ether leads to the formation of , and and the reaction rate is given by The rate of reaction is followed by increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether, i.e., If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants?

** 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.

** Question ** ** 4.7 ** What is the effect of temperature on the rate constant of a reaction? How can this effect of temperature on rate constant be represented quantitatively?

** 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.

** Question ** ** 4.10 ** In a reaction between A and B, the initial rate of reaction (r0) was measure for different initial concentrations of A and B as given below:

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

** Question ** ** 4.12 ** The reaction between A and B is first order with respect to A and zero order

with respect to B. Fill in the blanks in the following table:

Experiment | [A]/molL^{-1} | [B]/molL^{-1} | Initial rate/ mol L^{-1} min^{-1} |

I | 0.1 | 0.1 | 2*10^{-2} |

II | - | 0.2 | 4*10^{-2} |

III | 0.4 | 0.4 | - |

IV | 0.2 | 2*10^{-2} |

** 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)

** Question ** ** 4.14 ** The half-life for radioactive decay of is 5730 years. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample.

** 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:

Draw a graph between and t.

** 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:

What is the rate law ?

** 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:

Calculate the rate constant.

** 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

** Question ** ** 4.17 ** During nuclear explosion, one of the products is with half-life of 28.1 years. If of was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically.

** 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.

** Question ** ** 4.18 ** For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.

** 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.

Calculate the rate constant.

** 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:

Draw a graph between ln k and 1/T and calculate the values of A and

. Predict the rate constant at 30° and 50°C.

** 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 **

** Question ** ** 4.23 ** The rate constant for the decomposition of hydrocarbons is at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.

** 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)

** Question ** ** 4.24 ** Consider a certain reaction A Products with . Calculatethe concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol

** 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

** Question ** ** 4.25 ** Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with hours. What fraction of sample of sucrose remains after 8 hours ?

** 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 k=(4.51011s^{-1})e^{-28000K/T} . 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 =

** Question ** ** 4.29 ** The time required for 10% completion of a first order reaction at 298K is equal to that required for its 25% completion at 308K. If the value of A is . Calculate k at 318K and Ea.

** 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

** Question ** ** 4.30 ** The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.

** 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

## Topics and Sub-topics of Chemical Kinetics Class 12 NCERT Chapter 4

- 4.1 Rate of a Chemical Reaction
- 4.2 Factors Influencing Rate of a Reaction
- 4.3 Integrated Rate Equations
- 4.4 Temperature Dependence of the Rate of a Reaction
- 4.5 Collision Theory of Chemical Reactions

**NCERT Chapter-wise Solutions for Class 12 Chemistry**

Chapter 1 | The Solid State |

Chapter 2 | Solutions |

Chapter 3 | Electrochemistry |

Chapter 4 | Chemical Kinetics |

Chapter 5 | Surface chemistry |

Chapter 6 | General Principles and Processes of isolation of elements |

Chapter 7 | The P-block elements |

Chapter 8 | The d and f block elements |

Chapter 9 | Coordination compounds |

Chapter 10 | Haloalkanes and Haloarenes |

Chapter 11 | Alcohols, Phenols, and Ethers |

Chapter 12 | Aldehydes, Ketones and Carboxylic Acids |

Chapter 13 | Amines |

Chapter 14 | Biomolecules |

Chapter 15 | Polymers |

Chapter 16 | Chemistry in Everyday life |

### More About Class 12 Chemistry Chapter 4 Chemical Kinetics NCERT solutions

A total of 5 marks of questions will be asked in the Class 12 CBSE board exam of Chemistry from this chapter.

In this chapter, there are 30 questions in the exercise and 9 questions that are related to topics studied.

To clear doubts of students, the Chemical Kinetics NCERT solutions 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 the 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 chemistry chapter 4 PDF download, students can understand all the important concepts and practice questions well enough before their examination.

**NCERT Solutions for Class 12 Subject wise**

Which Topics Were Most Important For MHT CET Over The Past 5 Years? Read Here 6 min read Mar 18, 2022 Read More How Should You Prepare For The Physical Chemistry Questions In NEET And JEE Main? Read Here 7 min read Mar 05, 2022 Read More

- NCERT Solutions for Class 12 Biology
- NCERT solutions for class 12 Maths
- NCERT solutions for class 12 Chemistry
- NCERT Solutions for Class 12 Physics

### Benefits of NCERT solutions for Class 12 Chemistry Chapter 4 Chemical Kinetics

First, the easy steps given in the NCERT Class 12 Chemistry solutions chapter 4 will help you to understand the chapter easily.

The revision will be so easy 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.

**Also check NCERT Exemplar Class 12 Solutions**

- NCERT Exemplar Class 12th Maths Solutions
- NCERT Exemplar Class 12th Physics Solutions
- NCERT Exemplar Class 12th Chemistry Solutions
- NCERT Exemplar Class 12th Biology Solutions