NCERT Solutions for Class 12 Chemistry Chapter 3: Chemical Kinetics (NCERT 2026–27)

Q: How do you calculate the half-life of a first order reaction?

For a first order reaction, t1/2 = 0.693/k, where k is the rate constant. The half-life is constant and does not depend on the initial concentration of the reactant.

These Class 12 Chemistry Chapter 3 solutions cover Chemical Kinetics with step-by-step answers to every NCERT Intext Question and end-of-chapter Exercise. Every rate, order, half-life and Arrhenius numerical is solved fully with units and cross-checked against the NCERT answer key, so you can revise the speed of reactions, rate laws and the effect of temperature for the 2026–27 CBSE session with confidence.

Class: 12 Subject: Chemistry Chapter: 3 Name: Chemical Kinetics Exercises: 30 + 9 Intext Session: 2026–27

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Chapter overview
Key concepts
Important formulas
Intext questions — solutions
NCERT Exercises — solutions
Extra practice questions
MCQs & Assertion–Reason
Common mistakes
FAQs

Class 12 Chemistry Chapter 3 – Overview

Chemical kinetics is the branch of chemistry that studies how fast reactions occur and the factors that control their rates. While thermodynamics tells us whether a reaction is feasible, kinetics tells us its speed. The chapter defines the average and instantaneous rate of a reaction, develops the rate law and the ideas of order and molecularity, and derives integrated rate equations for zero- and first-order reactions along with the concept of half-life. It then explains the temperature dependence of rate through the Arrhenius equation and the role of activation energy and a catalyst, finishing with the collision theory of reaction rates. Mastering the numericals on order, rate constant, half-life and E_a is the key to scoring full marks here.

Key Concepts

Rate of reaction: change in concentration of a reactant or product per unit time (units mol L⁻¹ s⁻¹, or atm s⁻¹ for gases).

Rate law: Rate = k[A]^x[B]^y, determined experimentally; exponents need not equal stoichiometric coefficients.

Order of reaction: sum of the powers of concentration terms in the rate law (x + y). Can be 0, 1, 2, fractional.

Molecularity: number of species colliding in an elementary step; always a whole number (1–3), defined only for elementary reactions.

Rate constant (k): proportionality constant in the rate law; its units depend on overall order.

Half-life (t_1/2): time for the reactant concentration to fall to half its initial value.

Activation energy (E_a): minimum extra energy reactants must have to form the activated complex.

Pseudo first-order reaction: a higher-order reaction that behaves as first order because one reactant is in large excess (e.g. acid hydrolysis of an ester or cane sugar).

Important Formulas

Average rate: r_av = −Δ[R]/Δt = +Δ[P]/Δt (divide by the stoichiometric coefficient of each species).

Units of k: (mol L⁻¹)¹⁻ⁿ s⁻¹, where n = overall order — zero: mol L⁻¹ s⁻¹; first: s⁻¹; second: mol⁻¹ L s⁻¹.

Zero order: [R] = [R]₀ − kt; k = ([R]₀ − [R])/t; t_1/2 = [R]₀/2k.

First order: k = (2.303/t) log([R]₀/[R]); [R] = [R]₀e^−kt; t_1/2 = 0.693/k.

First order (gas, from pressures): k = (2.303/t) log[p_i/(2p_i − p_t)].

Arrhenius equation: k = A e^−E_a/RT; ln k = ln A − E_a/RT.

Two-temperature form: log(k₂/k₁) = (E_a/2.303R) · [(T₂ − T₁)/(T₁T₂)].

Constants: R = 8.314 J K⁻¹ mol⁻¹; log 2 = 0.301; ln x = 2.303 log x.

Intext Questions — Solutions

3.1 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.

SOLUTION r_av = −Δ[R]/Δt = −(0.02 − 0.03)/25 = 0.01/25 = 4 × 10⁻⁴ mol L⁻¹ min⁻¹. In seconds: 4 × 10⁻⁴ ÷ 60 = 6.66 × 10⁻⁶ mol L⁻¹ s⁻¹.

3.2 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?

SOLUTION Rate = −(1/2)·Δ[A]/Δt = −(1/2) × (0.4 − 0.5)/10 = (1/2) × (0.1/10). = (1/2) × 0.01 = 0.005 mol L⁻¹ min⁻¹ (i.e. 5 × 10⁻³ mol L⁻¹ min⁻¹).

3.3 For a reaction, A + B → Product; the rate law is given by, r = k[A]^1/2[B]². What is the order of the reaction?

SOLUTION Order = sum of powers = 1/2 + 2 = 2.5 (two and a half order).

3.4 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?

SOLUTION For X → Y, Rate = k[X]². New rate = k(3[X])² = 9k[X]². So the rate of formation of Y becomes 9 times the original rate.

3.5 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?

SOLUTION For first order: t = (2.303/k) log([R]₀/[R]) = (2.303 / 1.15 × 10⁻³) log(5/3). log(5/3) = log 1.667 = 0.2218. t = (2.303 / 1.15 × 10⁻³) × 0.2218 = 2002.6 × 0.2218 ≈ 444 s.

3.6 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.

SOLUTION For first order, t_1/2 = 0.693/k ⇒ k = 0.693/t_1/2. t_1/2 = 60 min = 60 × 60 = 3600 s. k = 0.693 / 3600 = 1.925 × 10⁻⁴ s⁻¹.

3.7 What will be the effect of temperature on rate constant?

SOLUTION The rate constant increases with temperature. As a rule of thumb, for every 10° rise in temperature the rate constant nearly doubles. Quantitatively this is described by the Arrhenius equation, k = A e^−E_a/RT: as T increases, the exponential term increases, so k increases (more molecules cross the activation energy barrier).

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

SOLUTION T₁ = 298 K, T₂ = 308 K, k₂/k₁ = 2. log(k₂/k₁) = (E_a/2.303R)[(T₂ − T₁)/(T₁T₂)]. log 2 = (E_a / 2.303 × 8.314)[(10)/(298 × 308)]. 0.301 = (E_a / 19.147) × (10 / 91784) = E_a × 5.689 × 10⁻⁶. E_a = 0.301 / 5.689 × 10⁻⁶ = 52911 J mol⁻¹ ≈ 52.9 kJ mol⁻¹.

3.9 The activation energy for the reaction 2 HI(g) → H₂ + I₂(g) is 209.5 kJ mol⁻¹ at 581 K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?

SOLUTION Fraction x = e^−E_a/RT; so ln x = −E_a/RT. ln x = −(209500 J mol⁻¹) / (8.314 × 581) = −209500 / 4830.4 = −43.37. log x = −43.37 / 2.303 = −18.832 ⇒ x = 10^−18.832 = 1.471 × 10⁻¹⁹.

NCERT Exercises — Solutions

3.1 From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants. (i) 3NO(g) → N₂O(g) Rate = k[NO]² (ii) H₂O₂(aq) + 3I⁻(aq) + 2H⁺ → 2H₂O(l) + I₃⁻ 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]

SOLUTION Units of k = (mol L⁻¹)¹⁻ⁿ s⁻¹ where n = order. (i) Order = 2. k units = (mol L⁻¹)⁻¹ s⁻¹ = mol⁻¹ L s⁻¹. (ii) Order = 1 + 1 = 2. k units = mol⁻¹ L s⁻¹. (iii) Order = 3/2. k units = (mol L⁻¹)^1−3/2 s⁻¹ = (mol L⁻¹)^−1/2 s⁻¹ = mol^−1/2 L^1/2 s⁻¹. (iv) Order = 1. k units = s⁻¹.

3.2 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.

SOLUTION Initial rate = k[A][B]² = (2.0 × 10⁻⁶)(0.1)(0.2)² = (2.0 × 10⁻⁶)(0.1)(0.04). = 2.0 × 10⁻⁶ × 4 × 10⁻³ = 8.0 × 10⁻⁹ mol L⁻¹ s⁻¹. After [A] falls to 0.06: A consumed = 0.1 − 0.06 = 0.04 mol L⁻¹. From 2A + B, B consumed = 0.04/2 = 0.02, so [B] = 0.2 − 0.02 = 0.18 mol L⁻¹. Rate = (2.0 × 10⁻⁶)(0.06)(0.18)² = (2.0 × 10⁻⁶)(0.06)(0.0324) = 3.89 × 10⁻⁹ mol L⁻¹ s⁻¹.

3.3 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?

SOLUTION Reaction: 2NH₃ → N₂ + 3H₂. For zero order, Rate = k = 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹. Rate = −(1/2)d[NH₃]/dt = d[N₂]/dt = (1/3)d[H₂]/dt. Rate of production of N₂ = k = 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹. Rate of production of H₂ = 3k = 3 × 2.5 × 10⁻⁴ = 7.5 × 10⁻⁴ mol L⁻¹ s⁻¹.

3.4 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(p_CH₃OCH₃)^3/2. If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants?

SOLUTION Since rate is expressed as change in pressure with time, units of rate = bar min⁻¹. From Rate = k(p)^3/2: k = Rate / p^3/2 = bar min⁻¹ / bar^3/2. Units of k = bar^−1/2 min⁻¹.

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

SOLUTION The rate of a chemical reaction depends on: (i) concentration of reactants (and pressure for gaseous reactants), (ii) temperature, (iii) presence of a catalyst, and (iv) for heterogeneous reactions, the surface area of reactants and the nature of reactants. Exposure to radiation/light can also affect certain reactions.

3.6 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?

SOLUTION Rate = k[A]². (i) Doubled: Rate’ = k(2[A])² = 4k[A]² ⇒ rate becomes 4 times. (ii) Halved: Rate” = k([A]/2)² = (1/4)k[A]² ⇒ rate becomes 1/4 (one-fourth) of the original.

3.7 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?

SOLUTION The rate constant increases with rise in temperature; generally k nearly doubles for every 10° rise. Quantitatively this is given by the Arrhenius equation: k = A e^−E_a/RT, where A is the pre-exponential (frequency) factor, E_a is activation energy, R the gas constant and T the absolute temperature. Taking logarithm: ln k = ln A − E_a/RT, and for two temperatures: log(k₂/k₁) = (E_a/2.303R)[(T₂ − T₁)/(T₁T₂)].

3.8 In a pseudo first order reaction in water, the following results were obtained:

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

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

SOLUTION r_av = −Δ[A]/Δt = −(0.17 − 0.31)/(60 − 30) = 0.14/30. = 4.67 × 10⁻³ mol L⁻¹ s⁻¹.

3.9 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?

SOLUTION (i) −d[R]/dt = Rate = k[A][B]². (ii) New rate = k[A](3[B])² = 9k[A][B]² ⇒ rate becomes 9 times. (iii) New rate = k(2[A])(2[B])² = k · 2[A] · 4[B]² = 8k[A][B]² ⇒ rate becomes 8 times.

3.10 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?

SOLUTION Let Rate = k[A]^x[B]^y. Order in B: Compare exp 1 and 2: [A] is constant (0.20) while [B] changes from 0.30 to 0.10 but rate is unchanged (5.07 × 10⁻⁵). Rate is independent of [B] ⇒ y = 0. Order in A: Since y = 0, Rate = k[A]^x. Compare exp 2 and 3: r₃/r₂ = (1.43 × 10⁻⁴)/(5.07 × 10⁻⁵) = 2.82 = (0.40/0.20)^x = 2^x. 2^x = 2.82 ⇒ x log 2 = log 2.82 ⇒ x = 0.45/0.301 = 1.5. Order with respect to A = 1.5; order with respect to B = 0 (overall order 1.5).

3.11 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.

SOLUTION Let Rate = k[A]^x[B]^y. Order in A: Compare I and IV ([B] constant at 0.1): r_IV/r_I = (2.40 × 10⁻²)/(6.0 × 10⁻³) = 4 = (0.4/0.1)^x = 4^x ⇒ x = 1. Order in B: Compare II and III ([A] constant at 0.3): r_III/r_II = (2.88 × 10⁻¹)/(7.2 × 10⁻²) = 4 = (0.4/0.2)^y = 2^y ⇒ y = 2. Rate law: Rate = k[A][B]² (overall order 3). Rate constant (using exp I): k = Rate/([A][B]²) = (6.0 × 10⁻³)/(0.1)(0.1)² = (6.0 × 10⁻³)/(1.0 × 10⁻³) = 6.0 mol⁻² L² min⁻¹ (i.e. 6.0 M⁻² min⁻¹).

3.12 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

SOLUTION Rate = k[A] (zero order in B). From exp I: k = Rate/[A] = (2.0 × 10⁻²)/0.1 = 0.2 min⁻¹. Exp II ([A]=?): [A] = Rate/k = (4.0 × 10⁻²)/0.2 = 0.2 mol L⁻¹. Exp III (rate = ?): Rate = k[A] = 0.2 × 0.4 = 8.0 × 10⁻² mol L⁻¹ min⁻¹. Exp IV ([A]=?): [A] = Rate/k = (2.0 × 10⁻²)/0.2 = 0.1 mol L⁻¹.

3.13 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

SOLUTION For first order, t_1/2 = 0.693/k. (i) t_1/2 = 0.693/200 = 3.47 × 10⁻³ s. (ii) t_1/2 = 0.693/2 = 0.35 min. (iii) t_1/2 = 0.693/4 = 0.173 years.

3.14 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.

SOLUTION k = 0.693/t_1/2 = 0.693/5730 = 1.209 × 10⁻⁴ year⁻¹. t = (2.303/k) log([R]₀/[R]) = (2.303/1.209 × 10⁻⁴) log(100/80). log(100/80) = log 1.25 = 0.0969. t = (2.303/1.209 × 10⁻⁴) × 0.0969 = 19049 × 0.0969 ≈ 1845 years.

3.15 The experimental data for decomposition of N₂O₅ [2N₂O₅ → 4NO₂ + O₂] in gas phase at 318 K are given below:

t/s	0	400	800	1200	1600	2000	2400	2800	3200
10²×[N₂O₅]/mol L⁻¹	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).

SOLUTION (i) Plot of [N₂O₅] vs t: a smooth curve that falls steeply at first and then flattens (exponential decay), characteristic of a first order reaction. (Drawn from the table above; no image needed.) (ii) Half-life from the data: [N₂O₅]₀ = 1.63 × 10⁻². Half = 0.815 × 10⁻², which is reached at about t = 1450 s. So t_1/2 ≈ 1450 s (read from the curve). (iii) Plot of log[N₂O₅] vs t: computing log values (e.g. log(1.63 × 10⁻²) = −1.79; log(0.35 × 10⁻²) = −2.46) gives a straight line of negative slope, confirming first order kinetics. (iv) Rate law: since log[N₂O₅] vs t is linear, Rate = k[N₂O₅] (first order). (v) Rate constant (take t = 1600 s): k = (2.303/t) log([R]₀/[R]) = (2.303/1600) log(1.63/0.78) = (2.303/1600) log 2.09 = (2.303/1600)(0.3201) = 4.61 × 10⁻⁴ s⁻¹ (slope method gives ≈ 4.8 × 10⁻⁴ s⁻¹). (vi) Half-life from k: t_1/2 = 0.693/k = 0.693/(4.61 × 10⁻⁴) ≈ 1504 s, which agrees closely with the value read from the graph in (ii).

3.16 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 1/16th value?

SOLUTION t = (2.303/k) log([R]₀/[R]) = (2.303/60) log(16). log 16 = log 2⁴ = 4 × 0.301 = 1.204. t = (2.303/60) × 1.204 = 0.03838 × 1.204 = 4.6 × 10⁻² s.

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

SOLUTION k = 0.693/t_1/2 = 0.693/28.1 = 2.466 × 10⁻² year⁻¹. First order: log([R]₀/[R]) = kt/2.303. After 10 years: log(1/[R]) = (2.466 × 10⁻² × 10)/2.303 = 0.1071 ⇒ [R] = antilog(−0.1071) = 0.7814 mg. After 60 years: log(1/[R]) = (2.466 × 10⁻² × 60)/2.303 = 0.6425 ⇒ [R] = antilog(−0.6425) = 0.227 mg.

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

SOLUTION For first order, t = (2.303/k) log([R]₀/[R]). 99% complete: [R] = 0.01[R]₀, so t₉₉ = (2.303/k) log(100/1) = (2.303/k) × 2 = 4.606/k. 90% complete: [R] = 0.10[R]₀, so t₉₀ = (2.303/k) log(100/10) = (2.303/k) × 1 = 2.303/k. Therefore t₉₉/t₉₀ = 4.606/2.303 = 2, i.e. t₉₉ = 2 t₉₀. Hence proved.

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

SOLUTION 30% decomposed ⇒ [R] = 70% of [R]₀, [R]₀/[R] = 100/70. k = (2.303/t) log(100/70) = (2.303/40) log 1.4286 = (2.303/40)(0.1549) = 8.918 × 10⁻³ min⁻¹. t_1/2 = 0.693/k = 0.693/(8.918 × 10⁻³) = 77.7 min.

3.20 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.

SOLUTION Reaction: (CH₃)₂CHN=NCH(CH₃)₂(g) → N₂(g) + C₆H₁₄(g). Let initial pressure p_i = 35.0; if x is the fall in pressure of reactant, total p_t = p_i + x, so x = p_t − p_i, and p_reactant = p_i − x = 2p_i − p_t. k = (2.303/t) log[p_i/(2p_i − p_t)]. At t = 360 s: 2p_i − p_t = 70 − 54 = 16. k = (2.303/360) log(35/16) = (2.303/360) log 2.1875 = (2.303/360)(0.3398) = 2.17 × 10⁻³ s⁻¹. At t = 720 s: 2p_i − p_t = 70 − 63 = 7. k = (2.303/720) log(35/7) = (2.303/720) log 5 = (2.303/720)(0.6990) = 2.235 × 10⁻³ s⁻¹. Average k = 2.20 × 10⁻³ s⁻¹.

3.21 The following data were obtained during the first order thermal decomposition of SO₂Cl₂ at a constant volume. SO₂Cl₂(g) → SO₂(g) + Cl₂(g)

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

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

SOLUTION p_i = 0.5 atm. p_SO₂Cl₂ = 2p_i − p_t = 1.0 − p_t. Rate constant (at t = 100 s, p_t = 0.6): p_reactant = 1.0 − 0.6 = 0.4. k = (2.303/100) log(0.5/0.4) = (2.303/100) log 1.25 = (2.303/100)(0.0969) = 2.23 × 10⁻³ s⁻¹. Rate at p_t = 0.65 atm: p_SO₂Cl₂ = 1.0 − 0.65 = 0.35 atm. Rate = k × p_SO₂Cl₂ = (2.23 × 10⁻³)(0.35) = 7.8 × 10⁻⁴ atm s⁻¹.

3.22 The rate constant for the decomposition of N₂O₅ at various temperatures is given below:

T/°C	0	20	40	60	80
10⁵×k/s⁻¹	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 30° and 50°C.

SOLUTION ln k vs 1/T is a straight line with slope = −E_a/R and intercept = ln A. Using the two end points (0°C = 273 K and 80°C = 353 K): At 273 K, k = 0.0787 × 10⁻⁵ = 7.87 × 10⁻⁷ s⁻¹; at 353 K, k = 2140 × 10⁻⁵ = 2.14 × 10⁻² s⁻¹. log(k₂/k₁) = (E_a/2.303R)[(T₂ − T₁)/(T₁T₂)]. log(2.14 × 10⁻² / 7.87 × 10⁻⁷) = log(2.72 × 10⁴) = 4.434. (T₂ − T₁)/(T₁T₂) = 80/(273 × 353) = 80/96369 = 8.30 × 10⁻⁴. E_a = (4.434 × 2.303 × 8.314)/(8.30 × 10⁻⁴) = 84.9/(8.30 × 10⁻⁴) ≈ 1.0 × 10⁵ J mol⁻¹ = 102 kJ mol⁻¹. A from k = A e^−E_a/RT at 273 K: A ≈ 3.9 × 10¹³ s⁻¹ (consistent with the line’s intercept). Predicted k: interpolating the ln k vs 1/T line gives k(30°C, 303 K) ≈ 7.6 × 10⁻⁵ s⁻¹ and k(50°C, 323 K) ≈ 7.6 × 10⁻⁴ s⁻¹ (each lies between the tabulated neighbouring values).

3.23 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.

SOLUTION k = A e^−E_a/RT ⇒ log A = log k + E_a/(2.303RT). E_a/(2.303RT) = 179900/(2.303 × 8.314 × 546) = 179900/10454 = 17.21. log k = log(2.418 × 10⁻⁵) = −4.616. log A = −4.616 + 17.21 = 12.59 ⇒ A = antilog(12.59) = 3.9 × 10¹² s⁻¹.

3.24 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.

SOLUTION Units of k (s⁻¹) indicate a first order reaction. log([R]₀/[R]) = kt/2.303 = (2.0 × 10⁻² × 100)/2.303 = 2/2.303 = 0.8684. [R]₀/[R] = antilog(0.8684) = 7.39 ⇒ [R] = 1.0/7.39 = 0.135 mol L⁻¹.

3.25 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?

SOLUTION k = 0.693/t_1/2 = 0.693/3.00 = 0.231 h⁻¹. log([R]₀/[R]) = kt/2.303 = (0.231 × 8)/2.303 = 1.848/2.303 = 0.8024. [R]₀/[R] = antilog(0.8024) = 6.345 ⇒ [R]/[R]₀ = 1/6.345 = 0.158. So about 0.158 (15.8%) of the sucrose remains.

3.26 The decomposition of hydrocarbon follows the equation k = (4.5 × 10¹¹ s⁻¹) e^−28000K/T. Calculate E_a.

SOLUTION Compare with k = A e^−E_a/RT: the exponent gives E_a/R = 28000 K. E_a = 28000 × R = 28000 × 8.314 = 232792 J mol⁻¹ = 232.79 kJ mol⁻¹.

3.27 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?

SOLUTION Compare log k = log A − E_a/(2.303RT) with the given equation: E_a/(2.303R) = 1.25 × 10⁴ K. E_a = 1.25 × 10⁴ × 2.303 × 8.314 = 239339 J mol⁻¹ = 239.339 kJ mol⁻¹. Temperature for t_1/2 = 256 min: t_1/2 = 256 min = 256 × 60 = 15360 s. k = 0.693/15360 = 4.51 × 10⁻⁵ s⁻¹. log k = log(4.51 × 10⁻⁵) = −4.346. So −4.346 = 14.34 − (1.25 × 10⁴/T). 1.25 × 10⁴/T = 14.34 + 4.346 = 18.686 ⇒ T = 1.25 × 10⁴/18.686 = 669 K (≈ 668.9 K).

3.28 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?

SOLUTION T₁ = 283 K, k₁ = 4.5 × 10³; k₂ = 1.5 × 10⁴; E_a = 60000 J mol⁻¹. log(k₂/k₁) = (E_a/2.303R)[(T₂ − T₁)/(T₁T₂)]. log(1.5 × 10⁴/4.5 × 10³) = log(3.333) = 0.5229. E_a/2.303R = 60000/(2.303 × 8.314) = 60000/19.147 = 3133.4. 0.5229 = 3133.4 × (T₂ − 283)/(283 T₂) ⇒ (T₂ − 283)/(283 T₂) = 1.6687 × 10⁻⁴. T₂ − 283 = 0.04722 T₂ ⇒ 0.95278 T₂ = 283 ⇒ T₂ = 297.0 K = 24°C.

3.29 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.

SOLUTION At 298 K (10% complete): t = (2.303/k₁) log(100/90) = (2.303/k₁)(0.04576). At 308 K (25% complete): t = (2.303/k₂) log(100/75) = (2.303/k₂)(0.1249). Since the two times are equal: 0.04576/k₁ = 0.1249/k₂ ⇒ k₂/k₁ = 0.1249/0.04576 = 2.73. E_a: log(k₂/k₁) = (E_a/2.303R)[(T₂ − T₁)/(T₁T₂)]. log 2.73 = 0.4362 = (E_a/19.147)[10/(298 × 308)] = (E_a/19.147)(1.0894 × 10⁻⁴). E_a = (0.4362 × 19.147)/(1.0894 × 10⁻⁴) = 8.353/1.0894 × 10⁻⁴ = 76673 J mol⁻¹ ≈ 76.75 kJ mol⁻¹. k at 318 K from k = A e^−E_a/RT: log k = log A − E_a/(2.303RT) = log(4 × 10¹⁰) − 76673/(2.303 × 8.314 × 318) = 10.602 − 76673/6089.4 = 10.602 − 12.591 = −1.989. k = antilog(−1.989) = 1.026 × 10⁻² s⁻¹ (≈ 0.9965 × 10⁻² s⁻¹ per the NCERT key, the small difference arising from rounding).

3.30 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.

SOLUTION k₂/k₁ = 4, T₁ = 293 K, T₂ = 313 K. log 4 = (E_a/2.303R)[(T₂ − T₁)/(T₁T₂)]. 0.6021 = (E_a/19.147)[20/(293 × 313)] = (E_a/19.147)(20/91709) = (E_a/19.147)(2.181 × 10⁻⁴). E_a = (0.6021 × 19.147)/(2.181 × 10⁻⁴) = 11.529/2.181 × 10⁻⁴ = 52860 J mol⁻¹ ≈ 52.8 kJ mol⁻¹.

Extra Practice Questions

Short Answer Type Questions

Q1. Distinguish between order and molecularity of a reaction.

ANSWEROrder is the sum of powers of concentration terms in the experimentally determined rate law; it can be zero, fractional or whole and applies to elementary and complex reactions. Molecularity is the number of species colliding in an elementary step; it is always a whole number (1–3) and is meaningless for an overall complex reaction.

Q2. Why can order of a reaction be fractional but molecularity cannot?

ANSWEROrder is an experimental quantity reflecting the rate-determining step of a possibly multi-step reaction, so it can come out fractional. Molecularity counts the actual particles colliding in a single elementary step, which must be a whole number; you cannot have a fraction of a molecule colliding.

Q3. Define activation energy. How does a catalyst change it?

ANSWERActivation energy (E_a) is the minimum extra energy that reactant molecules must possess (above their average energy) to form the activated complex and react. A catalyst provides an alternative pathway with a lower E_a, so a larger fraction of molecules can cross the barrier and the rate increases, without the catalyst being consumed.

Q4. What is a pseudo first-order reaction? Give one example.

ANSWERA reaction of higher true order that behaves as first order because one reactant is present in large excess, so its concentration stays effectively constant. Example: acid hydrolysis of ethyl acetate (or inversion of cane sugar), where water is in large excess, so Rate = k[ester].

Q5. Why does the rate of a reaction decrease with the passage of time?

ANSWERAs a reaction proceeds, the reactants are consumed and their concentration falls. Since rate depends on reactant concentration (Rate = k[A]^x…), the rate decreases steadily with time (except for zero-order reactions, where it stays constant until a reactant runs out).

Long Answer Type Questions

Q1. Derive the integrated rate equation for a first order reaction and show that its half-life is independent of initial concentration.

ANSWERFor R → P, Rate = −d[R]/dt = k[R]. Separating: d[R]/[R] = −k dt. Integrating: ln[R] = −kt + I. At t = 0, [R] = [R]₀, so I = ln[R]₀. Hence ln([R]/[R]₀) = −kt, or k = (2.303/t) log([R]₀/[R]).At half-life, [R] = [R]₀/2: k = (2.303/t_1/2) log 2 ⇒ t_1/2 = (2.303 × 0.301)/k = 0.693/k. Since [R]₀ cancels out, t_1/2 for a first order reaction is constant and independent of initial concentration.

Q2. Explain the Arrhenius equation and describe how E_a and A can be obtained graphically.

ANSWERThe Arrhenius equation, k = A e^−E_a/RT, links the rate constant to temperature; A is the frequency factor (related to collision frequency) and e^−E_a/RT is the fraction of molecules with energy ≥ E_a.Taking logarithm: ln k = ln A − E_a/RT. A plot of ln k (y) against 1/T (x) is a straight line of slope −E_a/R and intercept ln A. So E_a = −R × slope, and A = antilog/exp of the intercept. Increasing T or lowering E_a raises k and hence the reaction rate.

Q3. State the main postulates of collision theory and explain why it sometimes fails to predict rates accurately.

ANSWERCollision theory treats molecules as hard spheres that react only on colliding. The rate depends on the collision frequency Z and the fraction of collisions that are energetic enough: Rate = Z_AB e^−E_a/RT. Only “effective collisions” — those with energy ≥ threshold energy and correct orientation — lead to products.To account for orientation, a steric (probability) factor P is added: Rate = P Z_AB e^−E_a/RT. The theory predicts rates well for atoms and simple molecules but deviates for complex molecules, because it ignores the internal structure of molecules and treats them merely as hard spheres.

MCQs & Assertion–Reason

1. The unit of rate constant for a first order reaction is:

(a) mol L⁻¹ s⁻¹ (b) s⁻¹ (c) mol⁻¹ L s⁻¹ (d) mol L s⁻¹

2. For a zero order reaction, the rate of reaction is:

(a) proportional to [R] (b) proportional to [R]² (c) independent of concentration (d) proportional to √[R]

3. The half-life of a first order reaction is 0.693/k. This means t_1/2 is:

(a) directly proportional to [R]₀ (b) inversely proportional to [R]₀ (c) independent of [R]₀ (d) proportional to [R]₀²

4. If the rate law is Rate = k[A]^1/2[B]^3/2, the overall order is:

(a) 1 (b) 2 (c) 3 (d) 1/2

5. According to the Arrhenius equation, k increases when:

(a) T decreases (b) E_a increases (c) T increases or E_a decreases (d) A decreases

6. A catalyst increases the rate of a reaction by:

(a) increasing E_a (b) lowering E_a (c) raising temperature (d) changing ΔG

7. Molecularity of a reaction can never be:

(a) one (b) two (c) zero or fractional (d) three

8. Inversion of cane sugar in acid solution is an example of a:

(a) zero order reaction (b) pseudo first-order reaction (c) second order reaction (d) third order reaction

9. For a reaction Rate = k[A]², if [A] is reduced to half, the rate becomes:

(a) half (b) double (c) one-fourth (d) four times

10. The decomposition of NH₃ on a hot platinum surface at high pressure is:

(a) first order (b) second order (c) zero order (d) third order

Answer key: 1-(b), 2-(c), 3-(c), 4-(b), 5-(c), 6-(b), 7-(c), 8-(b), 9-(c), 10-(c).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: The half-life of a first order reaction is independent of the initial concentration.

Reason: For a first order reaction, t_1/2 = 0.693/k, in which initial concentration does not appear.

A-R 2. Assertion: Order of a reaction can be fractional.

Reason: Order is determined experimentally and reflects the rate-determining step of the mechanism.

A-R 3. Assertion: A catalyst changes the equilibrium constant of a reaction.

Reason: A catalyst lowers the activation energy of the forward and backward reactions equally.

A-R 4. Assertion: Rate constant increases with temperature.

Reason: With rise in temperature, the fraction of molecules having energy ≥ E_a increases.

A-R 5. Assertion: Molecularity of a reaction is always a whole number.

Reason: Molecularity is the number of reacting species colliding simultaneously in an elementary reaction.

Answer key: 1-(A), 2-(A), 3-(D), 4-(A), 5-(A).

Common Mistakes to Avoid

Watch out for these

Confusing order (experimental, can be fractional/zero) with molecularity (whole number, elementary steps only).
Forgetting to divide the rate by the stoichiometric coefficient when defining rate for species like 2A or 2N₂O₅.
Writing the rate law from the balanced equation — it must be found experimentally.
Mixing up ln and log (ln x = 2.303 log x) in first-order and Arrhenius formulas.
Not converting time and temperature units (minutes ↔ seconds, °C → K) before substituting.
Quoting the wrong units of k — always derive them from (mol L⁻¹)¹⁻ⁿ s⁻¹.
Thinking a catalyst changes ΔG or K_eq — it only speeds up attainment of equilibrium by lowering E_a.

How to score full marks in this chapter

Write the working step by step: state the formula, substitute with units, then evaluate. For order/rate-law problems, always show the ratio method (compare two experiments where one concentration is constant). For Arrhenius questions, decide first whether you need the single-temperature form (k = A e^−E_a/RT) or the two-temperature form, and use R = 8.314 J K⁻¹ mol⁻¹ with log 2 = 0.301. End every numerical with the correct unit — markers award a mark for it.

Frequently Asked Questions

What is Class 12 Chemistry Chapter 3 Chemical Kinetics about?

Chapter 3 studies how fast chemical reactions occur. It covers average and instantaneous rate, rate law, order and molecularity, integrated rate equations for zero- and first-order reactions, half-life, the Arrhenius equation and activation energy, the effect of a catalyst, and collision theory.

What is the difference between order and molecularity?

Order is the experimentally found sum of concentration powers in the rate law and can be zero, fractional or whole. Molecularity is the number of species colliding in an elementary step; it is always a whole number (1 to 3) and is defined only for elementary reactions.

How do you calculate the half-life of a first order reaction?

For a first order reaction, t_1/2 = 0.693/k, where k is the rate constant. The half-life is constant and does not depend on the initial concentration of the reactant.

Are these Class 12 Chemistry Chapter 3 solutions free?

Yes. All solutions are free, original and follow the official NCERT Chemistry textbook for the 2026-27 session, with every numerical solved step by step and verified against the NCERT answer key.

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