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?

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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

?

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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]

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= 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.

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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,

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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

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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 =

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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

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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) =

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=

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.,

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If the pressure is measured in bar and tidame 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

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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,

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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.

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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

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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,

=

=

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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 Aakas

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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:

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]

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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 =

=

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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 :

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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. Aakas

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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.

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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

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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 ?

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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: Aak

ash I

nstitu

te

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

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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.

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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, Aakas

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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.

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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

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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,

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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:

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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

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Slope of line =

According to Arrhenius equations,

Slope =

12.30 8.314

= 102.27

Again,

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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,

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= (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 :

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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)

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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,

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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 =

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= 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-

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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

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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,

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Activation Energy = 52.86 KJ/mo l

This is the required activation energy

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