Chemical Kinetics

34 solved questions from Class 12 Chemistry, each with the full working. Tap any question to see the step-by-step solution.

• 1 What is the primary chemical reaction or bonding mechanism highlighted in 'Chemical Kinetics' (Class 12)? medium MCQ Reaction Mechanism & Bonding
• 2 What factor affects the rate or equilibrium constant of reactions in 'Chemical Kinetics'? medium MCQ Chemical Kinetics & Factors
• 3 Which functional group or IUPAC nomenclature rule applies to compounds in 'Chemical Kinetics'? medium MCQ IUPAC Nomenclature & Structure
• 4 What is the major product obtained in the characteristic reaction of 'Chemical Kinetics'? medium MCQ Reaction Products & Selectivity
• 5 A reaction mechanism consists of three elementary steps: (1) A;xrightarrowfast; B, (2) B + C;xrightarrowslow; D, (3) D;xrightarrowfast; E. Which step is the rate-determining step? medium 2 marks Rate-determining step in a mechanism
• 6 A reaction follows the rate law r = k[A]^2. If the concentration of A is 0.10 M at a certain instant and the instantaneous rate of reaction is 0.020 M,s^-1, find the rate constant k. easy 1 mark Determination of rate constant from rate law
• 7 Consider the following mechanism for the reaction 2A → P: 1) A k_1I (slow) 2) I + A k_2P (fast) where I is an intermediate. (a) Write the rate law for the overall reaction using the steady‑state approximation for I. (b) Identify the rate‑determining step and explain why. (c) State the order of the reaction with respect to A. Provide a concise answer with reasoning. hard 5 marks Deriving rate law from mechanism (steady‑state)
• 8 Which of the following statements correctly describes the effect of a catalyst on the activation energy (E_a) of a reaction? medium MCQ 2 marks Catalyst effect on activation energy
• 9 The rate constant for a first‑order reaction is k_1=2.0×10^-3 s^-1 at 300 K and k_2=8.0×10^-3 s^-1 at 350 K. Using the Arrhenius equation, calculate the activation energy E_a (in kJ·mol^-1). hard 4 marks Activation energy from Arrhenius equation
• 10 Consider the following mechanism for the reaction 2A → P: 1. 2A xrightarrowk_1 I (slow) 2. I + A xrightarrowk_2 P (fast) hard 4 marks Mechanistic analysis using steady‑state approximation
• 11 Consider the following elementary mechanism for the reaction A + B → P: 1. A + B xrightarrowk_1 I (slow) 2. I xrightarrowk_2 P (fast) Use the steady‑state approximation for the intermediate I to derive the rate law for the overall reaction, and identify the rate‑determining step. medium 3 marks Steady‑state approximation in mechanism analysis
• 12 At an instant the concentration of reactant A in a reaction following the rate law r = k[A]^2 is 0.15 M. The instantaneous rate of reaction is 0.045 M,s^-1. Find the rate constant k (in M^-1,s^-1). easy 1 mark Rate law of an elementary reaction
• 13 Which of the following statements correctly describes the effect of increasing temperature on the rate constant k of a chemical reaction? medium MCQ 3 marks Temperature effect on rate constant (Arrhenius)
• 14 Consider the following mechanism for the reaction between A, B and C: 1. A + B [k_-1]k_1 I (fast equilibrium) 2. I + C k_2 P (slow) Using the pre‑equilibrium approximation, (a) derive the rate law for the overall reaction, (b) state the overall order with respect to each reactant, and (c) identify the rate‑determining step. hard 5 marks Pre‑equilibrium approximation and overall rate law
• 15 The half‑life of a first‑order reaction is measured to be 10 min. (a) Calculate the rate constant k in s^-1. (b) How much time (in minutes) will it take for the concentration of the reactant to fall to 25% of its initial value? easy 2 marks First‑order half‑life and integrated rate law
• 16 For a second‑order reaction 2A → products, the initial concentration of A is 0.050 M and the rate constant is k = 0.20 M^-1s^-1. Calculate the time (in seconds) required for the concentration of A to decrease to 0.020 M. medium 3 marks Second‑order integrated rate law
• 17 Two rate constants of a first‑order reaction are measured: k_1 = 1.5×10^-4 s^-1 at T_1 = 280,K and k_2 = 6.0×10^-4 s^-1 at T_2 = 330,K. Using the Arrhenius equation, calculate the activation energy E_a in kJ mol^-1. hard 5 marks Arrhenius activation energy
• 18 The half‑life of a first‑order reaction is found to be 12 min. Calculate the rate constant k in s^-1. easy 2 marks First‑order half‑life
• 19 Consider the mechanism for the reaction between X, Y and Z: 1. X + Y ⇌ I (fast equilibrium, equilibrium constant K) 2. I + Z xrightarrowk_2 P (slow) medium 4 marks Pre‑equilibrium rate law
• 20 Which of the following statements correctly describes the effect of a catalyst on the activation energy (E_a) of a reaction? medium MCQ 2 marks Effect of catalyst on activation energy
• 21 Two rate constants for a first‑order reaction are measured: k_1=2.5×10^-3 s^-1 at T_1=300 K and k_2=1.2×10^-2 s^-1 at T_2=350 K. Using the Arrhenius equation, calculate the activation energy E_a in kJ·mol^-1. medium 3 marks Arrhenius equation and activation energy
• 22 The half‑life of the first‑order decomposition of N_2O_5 at 298 K is 30 min. Calculate the rate constant k in s^-1. easy 1 mark First‑order half‑life calculation
• 23 For the reaction between A, B and C the following mechanism is proposed: 1. A + B [k_-1]k_1 I (fast equilibrium, K=k_1/(k_-1)) 2. I + C k_2 J (slow) 3. J k_3 P (fast) hard 4 marks Pre‑equilibrium & rate law derivation
• 24 Two rate constants of a first‑order reaction are measured as k_1=3.0×10^-4,s^-1 at T_1=300,K and k_2=1.2×10^-3,s^-1 at T_2=350,K. Using the Arrhenius equation, calculate the activation energy E_a (in kJ mol^-1) and the pre‑exponential factor A. hard 5 marks Arrhenius activation energy calculation
• 25 A reaction proceeds via the following mechanism: 1. displaystyle A+B xrightleftharpoons[k_-1]k_1 I (fast equilibrium) 2. displaystyle I+C xrightarrowk_2 P (slow) Derive the rate law in terms of [A], [B], [C], the equilibrium constant K and k_2. State the overall order of the reaction with respect to each reactant and identify the rate‑determining step. medium 3 marks Pre‑equilibrium rate law derivation
• 26 The half‑life of a first‑order reaction at 350 K is found to be 9 min. Calculate the rate constant k (in s^-1). easy 1 mark First‑order half‑life relation
• 27 Which of the following statements correctly describes the effect of raising the temperature on the rate constant k of a chemical reaction? medium MCQ 1 mark Temperature dependence of $k$ (Arrhenius)
• 28 For the following reaction mechanism: 1) A + B [k_-1]k_1 I (fast equilibrium, K_1=k_1/(k_-1)) 2) I + C [k_-2]k_2 J (fast equilibrium, K_2=k_2/(k_-2)) 3) J k_3 P (slow) hard 5 marks Rate law from a mechanism with pre‑equilibrium
• 29 The half‑life of the first‑order decomposition of H_2O_2 at 320 K is 7.5 min. Calculate the rate constant k in s^-1. easy 2 marks First‑order half‑life
• 30 The rate constant for a first‑order reaction is k_1=4.0×10^-4 s^-1 at 280 K and k_2=2.5×10^-3 s^-1 at 330 K. Using the Arrhenius equation, calculate the activation energy E_a in kJ mol^-1. medium 3 marks Determination of $E_{a}$ from two rate constants
• 31 The acid-catalysed hydrolysis of ethyl acetate, CH_3COOC_2H_5 + H_2O H^+ CH_3COOH + C_2H_5OH, is carried out in a large excess of water so that [H_2O] is essentially constant at 55.5 mol L^-1. Under these conditions the reaction follows pseudo-first-order kinetics with a measured rate constant k'=1.10×10^-3 s^-1 at 298 K. medium 4 marks Pseudo-first-order kinetics and conversion to true rate constant
• 32 For a reaction in dilute aqueous solution, the experimentally determined rate law is rate=k[A]^(3/2)[B]. If, at a particular instant, the concentration of A is tripled and the concentration of B is halved (all other conditions unchanged), by what factor does the instantaneous rate change compared to the original rate? medium MCQ 2 marks Effect of concentration changes on rate from the rate law
• 33 A zero-order decomposition of a drug A in a buffer solution follows the integrated rate law [A]=[A]_0-kt. The rate constant is k=4.0×10^-3 mol L^-1 s^-1 and the initial concentration is [A]_0=0.50 mol L^-1. easy 3 marks Zero-order integrated rate equation and its half-life
• 34 A dilute aqueous reaction has the rate law rate=k[A]^(3/2)[B]^2. If the concentration of A increases from 0.01 M to 0.04 M while [B] remains 0.02 M, by what factor does the instantaneous rate change compared to the original rate? easy MCQ 5 marks Effect of concentration on rate with fractional exponent