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.

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.