NCERT Solutions for Class 12 Science Chemistry Chapter 4 – Chemical Kinetics

Find comprehensive NCERT Solutions for Class 12 Science Chemistry Chapter 4 on Chemical Kinetics, featuring clear step-by-step explanations. These solutions have gained immense popularity among Chemistry students in Class 12 Science, serving as valuable resources for homework completion and exam preparation. Free access to all questions and answers from Chapter 4 of the NCERT Book for Class 12 Science Chemistry is provided here, ensuring a convenient and effective study aid for students.

Page No 98:

Question 4.1:

For the reaction R → P, the concentration of a reactant changes from 0.03 M to 0.02 M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.

ANSWER:

Average rate of reaction 

= 6.67 × 10⁻⁶ M s⁻¹

Page No 98:

Question 4.2:

In a reaction, 2A → Products, the concentration of A decreases from 0.5 mol L⁻¹ to 0.4 mol L⁻¹ in 10 minutes. Calculate the rate during this interval?

ANSWER:

Average rate 

= 0.005 mol L⁻¹ min⁻¹

= 5 × 10⁻³ M min⁻¹

Page No 103:

Question 4.3:

For a reaction, A + B → Product; the rate law is given by,. What is the order of the reaction?

ANSWER:

The order of the reaction

= 2.5

Page No 103:

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 reaction X → Y follows second order kinetics.

Therefore, the rate equation for this reaction will be:

Rate = k[X]² (1)

Let [X] = a mol L⁻¹, then equation (1) can be written as:

Rate₁ = k .(a)²

= ka²

If the concentration of X is increased to three times, then [X] = 3a mol L⁻¹

Now, the rate equation will be:

Rate = k (3a)²

= 9(ka²)

Hence, the rate of formation will increase by 9 times.

Page No 111:

Question 4.5:

A first order reaction has a rate constant 1.15 10⁻³ s⁻¹. How long will 5 g of this reactant take to reduce to 3 g?

ANSWER:

From the question, we can write down the following information:

Initial amount = 5 g

Final concentration = 3 g

Rate constant = 1.15 10⁻³ s⁻¹

We know that for a 1^st order reaction,

= 444.38 s

= 444 s (approx)

Page No 111:

Question 4.6:

Time required to decompose SO₂Cl₂ to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.

ANSWER:

We know that for a 1^st order reaction,

It is given that t_1/2 = 60 min

Page No 116:

Question 4.7:

What will be the effect of temperature on rate constant?

ANSWER:

The rate constant of a reaction is nearly doubled with a 10° rise in temperature. However, the exact dependence of the rate of a chemical reaction on temperature is given by Arrhenius equation,

Where,

A is the Arrhenius factor or the frequency factor

T is the temperature

R is the gas constant

E_a is the activation energy

Page No 116:

Question 4.8:

The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate E_a.

ANSWER:

It is given that T₁ = 298 K

∴T₂ = (298 + 10) K

= 308 K

We also know that the rate of the reaction doubles when temperature is increased by 10°.

Therefore, let us take the value of k₁ = k and that of k₂ = 2k

Also, R = 8.314 J K⁻¹ mol⁻¹

Now, substituting these values in the equation:

We get:

= 52897.78 J mol⁻¹

= 52.9 kJ mol⁻¹

Note: There is a slight variation in this answer and the one given in the NCERT textbook.

Page No 116:

Question 4.9:

The activation energy for the reaction

2HI_(g)→ H₂ + I_2(g)

is 209.5 kJ mol⁻¹ at 581K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?

ANSWER:

In the given case:

E_a = 209.5 kJ mol⁻¹ = 209500 J mol⁻¹

T = 581 K

R = 8.314 JK⁻¹ mol⁻¹

Now, the fraction of molecules of reactants having energy equal to or greater than activation energy is given as:

Page No 117:

Question 4.1:

From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.

(i) 3 NO(g) → N₂O(g) Rate = k[NO]²

(ii) H₂O₂ (aq) + 3 I⁻ (aq) + 2 H⁺ → 2 H₂O (l) +  Rate = k[H₂O₂][I⁻]

(iii) CH₃CHO(g) → CH₄(g) + CO(g) Rate = k [CH₃CHO]^3/2

(iv) C₂H₅Cl(g) → C₂H₄(g) + HCl(g) Rate = k [C₂H₅Cl]

ANSWER:

(i) Given rate = k [NO]²

Therefore, order of the reaction = 2

Dimension of 

(ii) Given rate = k [H₂O₂] [I⁻]

Therefore, order of the reaction = 2

Dimension of 

(iii) Given rate = k [CH₃CHO]^3/2

Therefore, order of reaction = 

Dimension of 

(iv) Given rate = k [C₂H₅Cl]

Therefore, order of the reaction = 1

Dimension of 

Page No 117:

Question 4.2:

For the reaction:

2A + B → A₂B

the rate = k[A][B]² with k = 2.0 × 10⁻⁶ mol⁻² L² s⁻¹. Calculate the initial rate of the reaction when [A] = 0.1 mol L⁻¹, [B] = 0.2 mol L⁻¹. Calculate the rate of reaction after [A] is reduced to 0.06 mol L⁻¹.

ANSWER:

The initial rate of the reaction is

Rate = k [A][B]²

= (2.0 × 10⁻⁶ mol⁻² L² s⁻¹) (0.1 mol L⁻¹) (0.2 mol L⁻¹)²

= 8.0 × 10⁻⁹ mol⁻² L² s⁻¹

When [A] is reduced from 0.1 mol L⁻¹ to 0.06 mol⁻¹, the concentration of A reacted = (0.1 − 0.06) mol L⁻¹ = 0.04 mol L⁻¹

Therefore, concentration of B reacted = 0.02 mol L⁻¹

Then, concentration of B available, [B] = (0.2 − 0.02) mol L⁻¹

= 0.18 mol L⁻¹

After [A] is reduced to 0.06 mol L⁻¹, the rate of the reaction is given by,

Rate = k [A][B]²

= (2.0 × 10⁻⁶ mol⁻² L² s⁻¹) (0.06 mol L⁻¹) (0.18 mol L⁻¹)²

= 3.89 mol L⁻¹ s⁻¹

Page No 117:

Question 4.3:

The decomposition of NH₃ on platinum surface is zero order reaction. What are the rates of production of N₂ and H₂ if k = 2.5 × 10⁻⁴ mol⁻¹ L s⁻¹?

ANSWER:

The decomposition of NH₃ on platinum surface is represented by the following equation.

Therefore,

However, it is given that the reaction is of zero order.

Therefore,

Therefore, the rate of production of N₂ is

And, the rate of production of H₂ is

= 7.5 × 10⁻⁴ mol L⁻¹ s⁻¹

Page No 117:

Question 4.4:

The decomposition of dimethyl ether leads to the formation of CH₄, H₂ and CO and the reaction rate is given by

Rate = k [CH₃OCH₃]^3/2

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 andtime in minutes, then what are the units of rate and rate constants?

ANSWER:

If pressure is measured in bar and time in minutes, then

Unit of rate = bar min⁻¹

Therefore, unit of rate constants

Page No 117:

Question 4.5:

Mention the factors that affect the rate of a chemical reaction.

ANSWER:

The factors that affect the rate of a reaction are as follows.

(i) Concentration of reactants (pressure in case of gases)

(ii) Temperature

(iii) Presence of a catalyst

Page No 118:

Question 4.6:

A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is

(i) doubled (ii) reduced to half?

ANSWER:

Letthe concentration of the reactant be [A] = a

Rate of reaction, R = k [A]²

= ka²

(i)If the concentration of the reactant is doubled, i.e. [A] = 2a, then the rate of the reaction would be

= 4ka²

= 4 R

Therefore, the rate of the reaction would increase by 4 times.

(ii) If the concentration of the reactant is reduced to half, i.e. , then the rate of the reaction would be

Therefore, the rate of the reaction would be reduced to

Page No 118:

Question 4.7:

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

ANSWER:

The rate constant is nearly doubled with a rise in temperature by 10° for a chemical reaction.

The temperature effect on the rate constant can be represented quantitatively by Arrhenius equation,

where, k is the rate constant,

A is the Arrhenius factor or the frequency factor,

R is the gas constant,

T is the temperature, and

E_a is the energy of activation for the reaction

Page No 118:

Question 4.8:

In a pseudo first order hydrolysis of ester in water, the following results were obtained:

t/s	0	30	60	90
[Ester]mol L−1	0.55	0.31	0.17	0.085

(i) Calculate the average rate of reaction between the time interval 30 to 60 seconds.

(ii) Calculate the pseudo first order rate constant for the hydrolysis of ester.

ANSWER:

(i) Average rate of reaction between the time interval, 30 to 60 seconds, 

= 4.67 × 10⁻³ mol L⁻¹ s⁻¹

(ii) For a pseudo first order reaction,

For t = 30 s, 

= 1.911 × 10⁻² s⁻¹

For t = 60 s, 

= 1.957 × 10⁻² s⁻¹

For t = 90 s, 

= 2.075 × 10⁻² s⁻¹

Then, average rate constant, 

Page No 118:

Question 4.9:

A reaction is first order in A and second order in B.

(i) Write the differential rate equation.

(ii) How is the rate affected on increasing the concentration of B three times?

(iii) How is the rate affected when the concentrations of both A and B are doubled?

ANSWER:

(i) The differential rate equation will be

(ii) If the concentration of B is increased three times, then

Therefore, the rate of reaction will increase 9 times.

(iii) When the concentrations of both A and B are doubled,

Therefore, the rate of reaction will increase 8 times.

Page No 118:

Question 4.10:

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

A/ mol L−1	0.20	0.20	0.40
B/ mol L−1	0.30	0.10	0.05
r0/ mol L−1 s−1	5.07 × 10−5	5.07 × 10−5	1.43 × 10−4

What is the order of the reaction with respect to A and B?

ANSWER:

Let the order of the reaction with respect to A be x and with respect to B be y.

Therefore,

Dividing equation (i) by (ii), we obtain

Dividing equation (iii) by (ii), we obtain

= 1.496

= 1.5 (approximately)

Hence, the order of the reaction with respect to A is 1.5 and with respect to B is zero.

Page No 118:

Question 4.11:

The following results have been obtained during the kinetic studies of the reaction:

2A + B → C + D

Experiment	A/ mol L−1	B/ mol L−1	Initial rate of formation of D/mol L−1 min−1
I	0.1	0.1	6.0 × 10−3
II	0.3	0.2	7.2 × 10−2
III	0.3	0.4	2.88 × 10−1
IV	0.4	0.1	2.40 × 10−2

Determine the rate law and the rate constant for the reaction.

ANSWER:

Let the order of the reaction with respect to A be x and with respect to B be y.

Therefore, rate of the reaction is given by,

According to the question,

Dividing equation (iv) by (i), we obtain

Dividing equation (iii) by (ii), we obtain

Therefore, the rate law is

Rate = k [A] [B]²

From experiment I, we obtain

= 6.0 L² mol⁻² min⁻¹

From experiment II, we obtain

= 6.0 L² mol⁻² min⁻¹

From experiment III, we obtain

= 6.0 L² mol⁻² min⁻¹

From experiment IV, we obtain

= 6.0 L² mol⁻² min⁻¹

Therefore, rate constant, k = 6.0 L² mol⁻² min⁻¹

Page No 118:

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/ mol L−1	B/ mol L−1	Initial rate/mol L−1 min−1
I	0.1	0.1	2.0 × 10−2
II	—	0.2	4.0 × 10−2
III	0.4	0.4	—
IV	—	0.2	2.0 × 10−2

ANSWER:

The given reaction is of the first order with respect to A and of zero order with respect to B.

Therefore, the rate of the reaction is given by,

Rate = k [A]¹ [B]⁰

⇒ Rate = k [A]

From experiment I, we obtain

2.0 × 10⁻² mol L⁻¹ min⁻¹ = k (0.1 mol L⁻¹)

⇒ k = 0.2 min⁻¹

From experiment II, we obtain

4.0 × 10⁻² mol L⁻¹ min⁻¹ = 0.2 min⁻¹ [A]

⇒ [A] = 0.2 mol L⁻¹

From experiment III, we obtain

Rate = 0.2 min⁻¹ × 0.4 mol L⁻¹

= 0.08 mol L⁻¹ min⁻¹

From experiment IV, we obtain

2.0 × 10⁻² mol L⁻¹ min⁻¹ = 0.2 min⁻¹ [A]

⇒ [A] = 0.1 mol L⁻¹

Page No 119:

Question 4.13:

Calculate the half-life of a first order reaction from their rate constants given below:

(i) 200 s⁻¹(ii) 2 min⁻¹(iii) 4 years⁻¹

ANSWER:

(i) Half life, 

= 3.47 ××10 ^-3 s (approximately)

(ii) Half life, 

= 0.35 min (approximately)

(iii) Half life, 

= 0.173 years (approximately)

Page No 119:

Question 4.14:

The half-life for radioactive decay of ¹⁴C 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:

Here, 

It is known that,

= 1845 years (approximately)

Hence, the age of the sample is 1845 years.

Page No 119:

Question 4.15:

The experimental data for decomposition of N₂O₅

in gas phase at 318K are given below:

t(s)	0	400	800	1200	1600	2000	2400	2800	3200
	1.63	1.36	1.14	0.93	0.78	0.64	0.53	0.43	0.35

(i) Plot [N₂O₅] against t.

(ii) Find the half-life period for the reaction.

(iii) Draw a graph between log [N₂O₅] and t.

(iv) What is the rate law?

(v) Calculate the rate constant.

(vi) Calculate the half-life period from k and compare it with (ii).

ANSWER:

(ii) Time corresponding to the concentration, is the half life. From the graph, the half life is obtained as 1450 s.

(iii)

t(s)
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

(iv) The given reaction is of the first order as the plot,  v/s t, is a straight line. Therefore, the rate law of the reaction is

(v) From the plot,  v/s t, we obtain

Again, slope of the line of the plot  v/s t is given by

.

Therefore, we obtain,

(vi) Half-life is given by,

This value, 1438 s, is very close to the value that was obtained from the graph.

Page No 119:

Question 4.16:

The rate constant for a first order reaction is 60 s⁻¹. How much time will it take to reduce the initial concentration of the reactant to its 1/16^th value?

ANSWER:

It is known that,

Hence, the required time is 4.6 × 10⁻² s.

Page No 119:

Question 4.17:

During nuclear explosion, one of the products is ⁹⁰Sr with half-life of 28.1 years. If 1μg of ⁹⁰Sr 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:

Here, 

It is known that,

Therefore, 0.7814 μg of ⁹⁰Sr will remain after 10 years.

Again,

Therefore, 0.2278 μg of ⁹⁰Sr will remain after 60 years.

Page No 119:

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:

For a first order reaction, the time required for 99% completion is

For a first order reaction, the time required for 90% completion is

Therefore, t₁ = 2t₂

Hence, the time required for 99% completion of a first order reaction is twice the time required for the completion of 90% of the reaction.

Page No 119:

Question 4.19:

A first order reaction takes 40 min for 30% decomposition. Calculate t_1/2.

ANSWER:

For a first order reaction,

Therefore, t_1/2 of the decomposition reaction is

= 77.7 min (approximately)

Page No 119:

Question 4.20:

For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained.

t (sec)	P(mm of Hg)
0	35.0
360	54.0
720	63.0

Calculate the rate constant.

ANSWER:

The decomposition of azoisopropane to hexane and nitrogen at 543 K is represented by the following equation.

After time, t, total pressure, 

= 2P₀ − P_t

For a first order reaction,

When t = 360 s, 

= 2.175 × 10⁻³ s⁻¹

When t = 720 s, 

= 2.235 × 10⁻³ s⁻¹

Hence, the average value of rate constant is

= 2.21 × 10⁻³ s⁻¹

Note: There is a slight variation in this answer and the one given in the NCERT textbook.

Page No 119:

Question 4.21:

The following data were obtained during the first order thermal decomposition of SO₂Cl₂ at a constant volume.

Experiment	Time/s−1	Total pressure/atm
1	0	0.5
2	100	0.6

Calculate the rate of the reaction when total pressure is 0.65 atm.

ANSWER:

The thermal decomposition of SO₂Cl₂ at a constant volume is represented by the following equation.

After time, t, total pressure, 

Therefore, 

= 2 P₀ − P_t

For a first order reaction,

When t = 100 s, 

= 2.231 × 10⁻³ s⁻¹

When P_t = 0.65 atm,

P₀ + p = 0.65

⇒ p = 0.65 − P₀

= 0.65 − 0.5

= 0.15 atm

Therefore, when the total pressure is 0.65 atm, pressure of SOCl₂ is

= P₀ − p

= 0.5 − 0.15

= 0.35 atm

Therefore, the rate of equation, when total pressure is 0.65 atm, is given by,

Rate = k()

= (2.23 × 10⁻³ s⁻¹) (0.35 atm)

= 7.8 × 10⁻⁴ atm s⁻¹

Page No 120:

Question 4.22:

The rate constant for the decomposition of N₂O₅at various temperatures is given below:

T/°C	0	20	40	60	80
	0.0787	1.70	25.7	178	2140

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

Predict the rate constant at 30º and 50ºC.

ANSWER:

From the given data, we obtain

T/°C	0	20	40	60	80
T/K	273	293	313	333	353
	3.66×10−3	3.41×10−3	3.19×10−3	3.0×10−3	2.83 ×10−3
	0.0787	1.70	25.7	178	2140
ln k	−7.147	− 4.075	−1.359	−0.577	3.063

Slope of the line,

According to Arrhenius equation,

Again,

When ,

Then, 

Again, when ,

Then, at ,

Page No 120:

Question 4.23:

The rate constant for the decomposition of hydrocarbons is 2.418 × 10⁻⁵ s⁻¹ at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.

ANSWER:

k = 2.418 × 10⁻⁵ s⁻¹

T = 546 K

E_a = 179.9 kJ mol⁻¹ = 179.9 × 10³ J mol⁻¹

According to the Arrhenius equation,

= (0.3835 − 5) + 17.2082

= 12.5917

Therefore, A = antilog (12.5917)

= 3.9 × 10¹² s⁻¹ (approximately)

Page No 120:

Question 4.24:

Consider a certain reaction A → Products with k = 2.0 × 10⁻² s⁻¹. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L⁻¹.

ANSWER:

k = 2.0 × 10⁻² s⁻¹

T = 100 s

[A]_o = 1.0 moL⁻¹

Since the unit of k is s⁻¹, the given reaction is a first order reaction.

Therefore, 

= 0.135 mol L⁻¹ (approximately)

Hence, the remaining concentration of A is 0.135 mol L⁻¹.

Page No 120:

Question 4.25:

Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t_1/2 = 3.00 hours. What fraction of sample of sucrose remains after 8 hours?

ANSWER:

For a first order reaction,

It is given that, t_1/2 = 3.00 hours

Therefore, 

= 0.231 h⁻¹

Then, 0.231 h⁻¹

Hence, the fraction of sample of sucrose that remains after 8 hours is 0.158.

Page No 120:

Question 4.26:

The decomposition of hydrocarbon follows the equation

k = (4.5 × 10¹¹ s⁻¹) e^{−28000 K/T}

Calculate E_a.

ANSWER:

The given equation is

k = (4.5 × 10¹¹ s⁻¹) e^{−28000 K/T} (i)

Arrhenius equation is given by,

 (ii)

From equation (i) and (ii), we obtain

= 8.314 J K⁻¹ mol⁻¹ × 28000 K

= 232792 J mol⁻¹

= 232.792 kJ mol⁻¹

Page No 120:

Question 4.27:

The rate constant for the first order decomposition of H₂O₂ is given by the following equation:

log k = 14.34 − 1.25 × 10⁴ K/T

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

ANSWER:

Arrhenius equation is given by,

The given equation is

From equation (i) and (ii), we obtain

= 1.25 × 10⁴ K × 2.303 × 8.314 J K⁻¹ mol⁻¹

= 239339.3 J mol⁻1 (approximately)

= 239.34 kJ mol⁻¹

Also, when t_1/2 = 256 minutes,

= 2.707 × 10⁻³ min⁻¹

= 4.51 × 10⁻⁵ s⁻¹

It is also given that, log k = 14.34 − 1.25 × 10⁴ K/T

= 668.95 K

= 669 K (approximately)

Page No 120:

Question 4.28:

The decomposition of A into product has value of k as 4.5 × 10³ s⁻¹ at 10°C and energy of activation 60 kJ mol⁻¹. At what temperature would k be 1.5 × 10⁴ s⁻¹?

ANSWER:

From Arrhenius equation, we obtain

Also, k₁ = 4.5 × 10³ s⁻¹

T₁ = 273 + 10 = 283 K

k₂ = 1.5 × 10⁴ s⁻¹

E_a = 60 kJ mol⁻¹ = 6.0 × 10⁴ J mol⁻¹

Then,

= 297 K

= 24°C

Hence, k would be 1.5 × 10⁴ s⁻¹ at 24°C.

Note: There is a slight variation in this answer and the one given in the NCERT textbook.

Page No 120:

Question 4.29:

The time required for 10% completion of a first order reaction at 298 K is

equal to that required for its 25% completion at 308 K. If the value of A is

4 × 10¹⁰ s⁻¹. Calculate k at 318 K and E_a.

ANSWER:

For a first order reaction,

At 298 K, 

At 308 K, 

According to the question,

From Arrhenius equation, we obtain

To calculate k at 318 K,

It is given that, 

Again, from Arrhenius equation, we obtain

Page No 120:

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 Arrhenius equation, we obtain

Hence, the required energy of activation is 52.86 kJmol⁻¹.