Question

NCERT Solutions Class 12 Chemistry Chapter 4 Chemical Kinetics is one of the most important chapters of chemistry to understand how chemical reactions take place. Our NCERT Solutions is a complete package for CBSE board exam preparations. NCERT Solutions Class 12 is based on the latest syllabus of the CBSE.

• Chemical Kinetics – the science of study of the rate at which chemical reaction occurs. The chemical kinetic establishes a relationship between many aspects of cosmology, geology, biology, engineering, and even psychology. Chemical kinetics also applies to the physical process and chemical reactions. Chemical kinematics also provides the study of the mechanism of a chemical process.
• Rate of a Chemical Reaction – the rate of a chemical reaction is calculated in the form of rates the product formed and reactant is consumed. The concentration of a substance is termed in the rates. It is easy to explain the rate at several molecules formed per unit of time.
• Half-life – the half-life is defined as the time taken by a chemical reaction to perform half of its chemical reaction. The half-life of any chemical reaction is not dependent on the initial amount of the reactants. Half-life is termed as when the half amount is left when some time of chemical reaction is already passed.
• Factors Influencing Rate of a Reaction – during of chemical reaction many factors affect it. Some of the factors influencing the rate of reaction are:
  • Reactant concentration – it seems that when we increase the concentration of one or more reactants will also increase the rate of reaction. This can be also seen that the collision of reactants is more when there is more number of reactant particles in the reaction process.
  • Physical state of the reactants and surface area – the rate of the reaction is always limited by the surface chemical reactants that are in contact with each other.

Our NCERT Solutions Class 12 Chemistry has discussed all the topics thoroughly so provide a basic understanding of all topics.

Answer 1

NCERT Solutions Class 12 Chemistry Chapter 4 Chemical Kinetics

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) + I⁻ 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 \(k = \frac{Rate}{[NO]^2}\)

(ii) Given rate = k [H₂O₂] [I⁻] 

Therefore, order of the reaction = 2

Dimension of \(k = \frac{Rate}{[H_2O_2][I^-]}\)

(iii) Given rate = k [CH₃CHO]^3/2

Therefore, order of reaction = 3/2

Dimension of  \(k = \frac{Rate}{[CH_3CHO]^\frac32}\)

(iv) Given rate = k [C₂H₅Cl] Therefore, order of the reaction = 1

Dimension of \(k = \frac{Rate}{[C_2H_2Cl]}\)

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 \( \frac12\times0.04\,mol\,L^{-1}\) = 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

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

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.,

Rate = \(k\left(p_{_{CH_3OCH_3}}\right)^{\frac32}\)

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

Rate = \(k\left(p_{_{CH_3OCH_3}}\right)^{\frac32}\) 

⇒ \(k = \frac{Rate}{\left(p_{_{CH_3OCH_3}}\right)^{\frac32}}\)

Therefore, unit of rate constants \((k) = \frac{bar\,min^{-1}}{bar^{\frac32}}\)

= \(bar^{\frac{-1}2} min^{-1}\) 

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

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:

Let the 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

R' = k(2a)²

= 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. [A] = \(\frac12a\), then the rate of the reaction would be

Therefore, the rate of the reaction would be reduced to \(\frac14^{th}.\)

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,

\(k=Ae^{-E_a/RT}\)

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

Answer 2

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, 

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.

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.

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,

Rate = k[A]^x [B]^y

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

⇒ \(k =\frac{Rate }{[A] [B]^2}\)

Answer 3

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

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

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,

\(t_{\frac12}=\frac{0.693}{k}\)

\(=\frac{0.693}{200\,s^{-1}}\)

= 3.47 x 10^-3 s (approximately)

(ii) Half life,

\(t_{\frac12}=\frac{0.693}{k}\)

\(=\frac{0.693}{2\,min^{-1}}\) 

= 0.35 min (approximately)

(iii) Half life,

\(t_{\frac12}=\frac{0.693}{k}\)

\(=\frac{0.693}{4\,years^{-1}}\)

= 0.173 years (approximately)

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,

\(k=\frac{0.693}{t_{\frac12}}\)

\(=\frac{0.693}{5730}years^{-1}\)

It is known that,

= 1845 years (approximately) 

Hence, the age of the sample is 1845 years.

15. The experimental data for decomposition of N₂O₅

[2N₂O₅ → 4NO₂ + O₂]

in gas phase at 318K are given below:

(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:

(i)

(ii) Time corresponding to the concentration, \(\frac{1.630\times10^2}{2}mol L^{-1}= 81.5\,molL^{-1}\) is the half life. From the graph, the half life is obtained as 1450 s.h.

Answer 4

(iii) 

t(s)	102 × [N2​O5​] mol L−1	log [N2​O5​]
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, log[N₂​O₅​] v/s t, is a straight line.

Therefore, the rate law of the reaction is

Rate = k[N₂​O₅​]

(v) From the plot, log[N₂​O_5​] v/s t, we obtain

Again, slope of the line of the plot log[N₂​O₅​] v/s t is given by \(-\frac{k}{2.303}.\)

Therefore, we obtain,

(vi) Half-life is given by,

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

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.

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.

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.

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)

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

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, P_t​ = (P₀​ – p) + p + p

⇒ P_t ​= P₀​ + p

⇒ p = P_t​ – P₀

​Therefore, P₀​ – p = P₀​ – (P_t​ – P₀​)

= 2P₀ ​– P_t​

For the 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

Answer 5

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

SO₂Cl₂(g) → SO₂ + Cl₂(g)

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, P_t​ = (P₀​ – p) + p + p

⇒ P_t ​= P₀​ + p

⇒ p = P_t​ – P₀

​Therefore, P₀​ – p = P₀​ – (P_t​ – P₀​)

= 2P₀ ​– 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

\(p_{_{SOCl_2}}\) = 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(\(p_{_{SOCl_2}}\))

= (2.23 × 10⁻³ s⁻¹) (0.35 atm) 

= 7.8 × 10⁻⁴ atm s⁻¹

22. The rate constant for the decomposition of hydrocarbons is 2.418 × 10^–5s^–1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of the 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)

23. 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, \(k = \frac{2.303}{t}log\frac{[A]_o}{[A]}\)

= 0.135 mol L⁻¹ (approximately) 

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

24. 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, \(k = \frac{0.693}{t_\frac12}\)

\(=\frac{0.693}{3}h^{-1}\)

= 0.231 h⁻¹

Then, 0.231 h⁻¹ \(= \frac{2.303}{8h}log\frac{[R]_0}{[R]}\)

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

25. 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,

k = Ae^-Ea/RT    .....(ii)

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

\(\frac{E_a}{RT}=\frac{28000\,K}{T}\)

⇒ E_a ​= R × 28000K

= 8.314 J K⁻¹ mol⁻¹ × 28000 K 

= 232792 J mol⁻¹ 

= 232.792 kJ mol⁻¹

26. 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,

\(k = \frac{0.693}{t_\frac12}\)

\(= \frac{0.693}{256}\)

= 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)

Answer 6

27. 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 × 10⁴s^–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.

28. 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 4 × 10¹⁰s ^–1. Calculate k at 318K 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, A = 4 x 10¹⁰ s^-1, T = 318 K

Again, from Arrhenius equation, we obtain

29. 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.