NCERT Solutions for Class 12 Chemistry Chapter 3: Chemical Kinetics – Free PDF Download

Chapter 3 covers chemical kinetics — rate of reaction, rate laws and order of reaction, integrated rate equations for zero- and first-order reactions, half-life, the effect of temperature via the Arrhenius equation, and factors affecting reaction rate. All 29 exercises (3.1–3.29) remain current under the 2026-27 syllabus, with no separate “Additional Exercises” section. Each numeric answer below is verified before publishing.

Last Updated: September 23, 2026

NCERT Exercise 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].
Ans: (i) Order=2; k units=mol⁻¹Ls⁻¹. (ii) Order=1+1=2; k units=mol⁻¹Ls⁻¹. (iii) Order=3/2; k units=mol^−1/2L^1/2s⁻¹. (iv) Order=1; k units=s⁻¹.

3.2 For the reaction 2A+B→A₂B, the rate=k[A][B]² with k=2.0×10⁻⁶mol⁻²L²s⁻¹. Calculate the initial rate when [A]=0.1mol L⁻¹, [B]=0.2mol L⁻¹. Calculate the rate of reaction after [A] is reduced to 0.06mol L⁻¹.
Ans: Initial rate=k[A][B]²=2.0×10⁻⁶×0.1×(0.2)²=8.0×10⁻⁹mol L⁻¹s⁻¹. Since A and B react in a 2:1 ratio, when [A] falls by 0.04 (to 0.06), [B] falls by half that, 0.02, to 0.18mol L⁻¹. New rate=2.0×10⁻⁶×0.06×(0.18)²≈3.89×10⁻⁹mol L⁻¹s⁻¹.

3.3 The decomposition of NH₃ on a platinum surface is zero order. If k=2.5×10⁻⁴mol L⁻¹s⁻¹, what are the rates of production of N₂ and H₂?
Ans: For 2NH₃→N₂+3H₂ (zero order), rate of reaction=k=2.5×10⁻⁴mol L⁻¹s⁻¹. Since 1 mol N₂ forms per reaction event, d[N₂]/dt=2.5×10⁻⁴mol L⁻¹s⁻¹; since 3 mol H₂ forms per event, d[H₂]/dt=3×2.5×10⁻⁴=7.5×10⁻⁴mol L⁻¹s⁻¹.

3.4 The decomposition of dimethyl ether leads to CH₄, H₂ and CO, with rate=k[CH₃OCH₃]^3/2. If pressure is measured in bar and time in minutes, what are the units of rate and rate constant?
Ans: Order=3/2. Rate units=bar min⁻¹. Rate constant units=(bar min⁻¹)/(bar^3/2)=bar^−1/2min⁻¹.

3.5 Mention the factors that affect the rate of a chemical reaction.
Ans: Reaction rate is affected by: the concentration of reactants (higher concentration generally increases rate), temperature (rate increases with temperature since more molecules gain the activation energy needed to react), the presence of a catalyst (which provides an alternate, lower-activation-energy pathway), the nature of the reactants (e.g. ionic reactions are typically faster than covalent-bond-breaking reactions), and the surface area of reactants in heterogeneous reactions (a finely powdered solid reacts faster than a lump).

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?
Ans: Since rate=k[X]²: (i) doubling [X] makes the rate 4 times the original (2²=4). (ii) halving [X] makes the rate one-quarter the original ((1/2)²=1/4).

3.7 What is the effect of temperature on the rate constant of a reaction? How can this temperature effect on the rate constant be represented quantitatively?
Ans: The rate constant of a reaction increases with temperature — as a rough rule, it roughly doubles for every 10°C rise for many reactions. This is represented quantitatively by the Arrhenius equation, k=Ae^−E_a/RT, where A is the pre-exponential (frequency) factor, E_a is the activation energy, R is the gas constant, and T is the absolute temperature — higher T makes the exponential term (and hence k) larger.

3.8 A pseudo first order reaction (ester hydrolysis) was followed by measuring ester concentration at intervals: t(s)=0, 30, 60, 90; [Ester](mol L⁻¹)=0.55, 0.31, 0.17, 0.085. Calculate the average rate of reaction between t=30s and t=60s, and the pseudo first order rate constant for the hydrolysis.
Ans: Average rate (30–60s)=(0.31−0.17)/30≈4.67×10⁻³mol L⁻¹s⁻¹. Using k=(1/t)ln([Ester]₀/[Ester]_t) for each interval: k₍₀₋₃₀₎≈1.911×10⁻², k_(30−60)≈2.003×10⁻², k_(60−90)≈2.310×10⁻²s⁻¹. Average pseudo first order k≈2.07×10⁻²s⁻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?
Ans: (i) Rate=−d[A]/dt=k[A][B]². (ii) Tripling [B] makes the rate 9 times the original (3²=9). (iii) Doubling both [A] and [B] makes the rate 8 times the original (2×2²=8).

3.10 In a reaction between A and B, the initial rate was measured for different concentrations: Expt 1, [A]=0.20M, [B]=0.30M, rate=5.07×10⁻⁵; Expt 2, [A]=0.20M, [B]=0.10M, rate=5.07×10⁻⁵; Expt 3, [A]=0.40M, [B]=0.05M, rate=1.43×10⁻⁴ (mol L⁻¹s⁻¹). What is the order of the reaction with respect to A and B?
Ans: Comparing Expt 1 and 2, changing [B] (0.30→0.10) with [A] constant does not change the rate, so order with respect to B=0. Comparing Expt 2 and 3, doubling [A] (0.20→0.40) increases the rate by a factor of 1.43×10⁻⁴/5.07×10⁻⁵≈2.82≈2^1.5, so order with respect to A=1.5.

3.11 The following results were obtained for the reaction 2A+B→C+D: Expt I, [A]=0.1,[B]=0.1, rate=6.0×10⁻³; Expt II, [A]=0.3,[B]=0.2, rate=7.2×10⁻²; Expt III, [A]=0.3,[B]=0.4, rate=2.88×10⁻¹; Expt IV, [A]=0.4,[B]=0.1, rate=2.4×10⁻² (mol L⁻¹min⁻¹). Determine the rate law and the rate constant.
Ans: Comparing II and III ([A] constant, [B] doubled): rate increases 4×, so order in B=2. Comparing I and IV ([B] constant, [A] quadrupled): rate increases 4×, so order in A=1. Rate law: Rate=k[A][B]². From Expt I: k=6.0×10⁻³/(0.1×0.1²)=6.0L²mol⁻²min⁻¹ (confirmed consistent across all four experiments).

3.12 The following data were obtained for a reaction that is first order in A and zero order in B: Expt 1, [A]=0.1M,[B]=0.1M,rate=2.0×10⁻²; Expt 2, [B]=0.2M, rate=4.0×10⁻²; Expt 3, [A]=0.4M,[B]=0.4M, rate=?; Expt 4, [B]=0.2M, rate=2.0×10⁻² (mol L⁻¹min⁻¹). Complete the table.
Ans: Rate=k[A] (B has no effect). From Expt 1: k=2.0×10⁻²/0.1=0.2min⁻¹. Expt 2: 4.0×10⁻²=0.2×[A] ⇒ [A]=0.2M. Expt 3: rate=0.2×0.4=0.08mol L⁻¹min⁻¹. Expt 4: 2.0×10⁻²=0.2×[A] ⇒ [A]=0.1M.

3.13 Calculate the half-lives of first-order reactions with rate constants: (i) 200s⁻¹ (ii) 2min⁻¹ (iii) 4year⁻¹.
Ans: Using t_1/2=0.693/k: (i) 0.693/200≈3.47×10⁻³s. (ii) 0.693/2≈0.35min. (iii) 0.693/4≈0.173year.

3.14 The half-life for radioactive decay of ¹⁴₆C 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.
Ans: k=0.693/5730≈1.209×10⁻⁴year⁻¹. Using t=(1/k)ln(100/80): t≈1845 years.

3.15 The decomposition of N₂O₅ at 318K was followed by monitoring [N₂O₅] (×10²mol L⁻¹) at intervals up to 3200s (data: 1.63, 1.36, 1.14, 0.93, 0.78, 0.64, 0.53, 0.43, 0.35 at t=0,400,…,3200). Confirm the reaction is first order and calculate the rate constant and half-life.
Ans: Applying k=(1/t)ln([N₂O₅]₀/[N₂O₅]_t) at each interval gives a near-constant k (ranging ≈4.5–4.8×10⁻⁴s⁻¹), confirming first order. Average k≈4.7×10⁻⁴s⁻¹, giving t_1/2=0.693/k≈1470s.

3.16 The rate constant for a first-order reaction is 60s⁻¹. How much time will it take to reduce the initial concentration to its 1/16th value?
Ans: Using t=(1/k)ln(16): t=(1/60)×ln(16)≈4.62×10⁻²s.

3.17 ⁹⁰Sr has a half-life of 28.1 years. Calculate the amount remaining after (i) 10 years (ii) 60 years, if the initial amount was 1μg.
Ans: k=0.693/28.1≈2.466×10⁻²year⁻¹. (i) N=1×e^−k×10≈0.781μg. (ii) N=1×e^−k×60≈0.228μg.

3.18 Show that for a first-order reaction, the time required for 99% completion is twice the time required for 90% completion.
Ans: t=(1/k)ln(100/(100−x)). For 90% completion (10% remains): t₉₀=(1/k)ln(10)=2.303/k. For 99% completion (1% remains): t₉₉=(1/k)ln(100)=4.606/k. So t₉₉/t₉₀=4.606/2.303=2, confirming t₉₉=2×t₉₀.

3.19 A first-order reaction takes 40 minutes for 30% decomposition. Calculate its rate constant and half-life.
Ans: k=(1/40)ln(100/70)≈8.92×10⁻³min⁻¹. t_1/2=0.693/k≈77.7min.

3.20 The decomposition of azoisopropane, (CH₃)₂CHN=NCH(CH₃)₂(g)→N₂(g)+C₆H₁₄(g), was monitored via total pressure at 543K: t(s)=0,360,720, P(mmHg)=35.0,54.0,63.0. Calculate the rate constant.
Ans: Since 1 mol reactant gives 2 mol product, P_reactant=2P₀−P_total. At t=360s: P_A=70−54=16.0mmHg, k≈2.17×10⁻³s⁻¹. At t=720s: P_A=70−63=7.0mmHg, k≈2.24×10⁻³s⁻¹. Average k≈2.21×10⁻³s⁻¹.

3.21 The first-order thermal decomposition of SO₂Cl₂(g)→SO₂(g)+Cl₂(g) at constant volume gave: Expt 1, t=0s, total pressure=0.5atm; Expt 2, t=100s, total pressure=0.6atm. Calculate the rate of the reaction when total pressure is 0.65atm.
Ans: Since P_SO2Cl2=P₀−x and P_total=P₀+x, at t=100s: x=0.1, P_SO2Cl2=0.4atm, giving k=(1/100)ln(0.5/0.4)≈2.23×10⁻³s⁻¹. At total pressure=0.65atm: x=0.15, P_SO2Cl2=0.35atm, so rate=k×P_SO2Cl2=2.23×10⁻³×0.35≈7.81×10⁻⁴atm s⁻¹.

3.22 The rate constant for the decomposition of a hydrocarbon is 2.418×10⁻⁵s⁻¹ at 546K. If the energy of activation is 179.9kJ/mol, what is the value of the pre-exponential factor?
Ans: Using k=Ae^−E_a/RT: A=k/e^−E_a/RT=2.418×10⁻⁵/e^{−179900/(8.314×546)}≈3.93×10¹²s⁻¹.

3.23 The rate constant for a first-order reaction A→Products is 2.0×10⁻²s⁻¹. Calculate the concentration of A remaining after 100s if the initial concentration was 1.0mol L⁻¹.
Ans: [A]_t=[A]₀e^−kt=1.0×e^{−2.0×10−2×100}=1.0×e⁻²≈0.135mol L⁻¹.

3.24 Sucrose decomposes in acid solution into glucose and fructose following first-order kinetics, with t_1/2=3.00 hours. What fraction of the sample of sucrose remains after 8 hours?
Ans: k=0.693/3.00≈0.231hour⁻¹. Fraction remaining=e^−kt=e^−0.231×8≈0.158 (about 15.8%).

3.25 The decomposition of a hydrocarbon follows k=4.5×10¹¹e^−28000K/Ts⁻¹. Calculate its activation energy.
Ans: Comparing with the Arrhenius equation, E_a/R=28000K, so E_a=28000×8.314≈232.8kJ/mol.

3.26 The rate constant for the first-order decomposition of H₂O₂ is given by log k=14.34−1.25×10⁴K/T. Calculate the activation energy and the temperature at which its half-life is 256 minutes.
Ans: Since log k=log A−E_a/(2.303RT), E_a=1.25×10⁴×2.303×8.314≈239.3kJ/mol. For t_1/2=256min=15360s: k=0.693/15360≈4.51×10⁻⁵s⁻¹, log k≈−4.346. Solving 14.34−1.25×10⁴/T=−4.346 gives T≈669K.

3.27 The decomposition of A into products has a rate constant of 4.5×10³s⁻¹ at 10°C, and the energy of activation is 60kJ/mol. At what temperature would k be 1.5×10⁴s⁻¹?
Ans: Using ln(k₂/k₁)=−(E_a/R)(1/T₂−1/T₁) with T₁=283K: solving gives T₂≈297K (24°C).

3.28 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 A=4×10¹⁰s⁻¹, calculate k at 318K and the energy of activation.
Ans: Since the times are equal, k₃₀₈/k₂₉₈=ln(100/75)/ln(100/90)≈2.73. Using the Arrhenius two-point equation: E_a≈76.6kJ/mol. Then k₃₁₈=Ae^{−E_a/(R×318)}≈1.03×10⁻²s⁻¹.

3.29 The rate of a reaction quadruples when the temperature changes from 293K to 313K. Calculate the energy of activation, assuming it does not change with temperature.
Ans: Using ln(4)=(E_a/R)(1/293−1/313): E_a≈52.9kJ/mol.

Frequently Asked Questions

What is the difference between the rate of reaction and the rate constant?
The rate of reaction is the change in concentration of a reactant or product per unit time at a given instant, and generally depends on the current concentrations of the reactants. The rate constant (k) is a fixed proportionality constant (at a given temperature) in the rate law connecting rate to concentrations — it does not depend on concentration, only on temperature and the nature of the reaction.

Why does the rate constant increase with temperature according to the Arrhenius equation?
The Arrhenius equation, k=Ae^−E_a/RT, shows that as temperature T rises, the negative exponent −E_a/RT becomes less negative, making e^−E_a/RT larger. Physically, higher temperature means more reactant molecules possess kinetic energy exceeding the activation energy E_a, so a larger fraction of collisions are effective, increasing k.

Written by Satish

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