NCERT Solutions for Class 12 Chemistry Chapter 4 Chemical

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


(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


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


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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Answer :

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

= 0.693/2

= 0.35 min (approx)



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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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



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



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,


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.

Again, Aakas

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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,


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,


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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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NCERT Solutions for Class 12 Chemistry Chapter 4 Chemical