CBSE · Class 12 · Chemistry

Class 12 Chemical Kinetics Mock Test (CBSE)

3 practice questions from Chemical Kinetics, in the style of your board paper, each with a worked answer and the marks for every step. Then take a full chapter test on paper, snap the pages and get every step marked.

Topics in Chemical Kinetics

  • Rate of a chemical reaction
  • Factors influencing rate of a reaction
  • Rate law and rate constant
  • Order and molecularity of a reaction
  • Integrated rate equations
  • Zero order reactions
  • First order reactions
  • Half-life of a reaction
  • Pseudo first order reactions
  • Temperature dependence and Arrhenius equation
  • Collision theory

Chemical Kinetics practice questions with answers

  1. Q1Multiple choice1 mark
    Which of the following statements is correct regarding the effect of a catalyst on a chemical reaction?
    1. (a)It lowers the activation energy of the forward and backward reactions equally.
    2. (b)It increases the activation energy of the reaction.
    3. (c)It changes the Gibbs free energy of the reaction.
    4. (d)It alters the equilibrium constant of the reaction.
    Show answer and step marks
    1. Correct option: (a) It lowers the activation energy of the forward and backward reactions equally. Recognize that a catalyst provides an alternative reaction pathway of lower activation energy, affecting both forward and backward reactions equally without changing thermodynamics or equilibrium constant, corresponding to option (a).+1
    Answer: (a)
  2. Q2Multiple choice1 mark
    For the gas-phase reaction 2N2O5→4NO2+O22N_2O_5 \rightarrow 4NO_2 + O_2, the rate of formation of NO2NO_2 is 2.8×10−3 M s−12.8 \times 10^{-3} \, \text{M s}^{-1}. What is the rate of disappearance of N2O5N_2O_5?
    1. (a)1.12×10−2 M s−11.12 \times 10^{-2} \, \text{M s}^{-1}
    2. (b)5.6×10−3 M s−15.6 \times 10^{-3} \, \text{M s}^{-1}
    3. (c)1.4×10−3 M s−11.4 \times 10^{-3} \, \text{M s}^{-1}
    4. (d)2.8×10−3 M s−12.8 \times 10^{-3} \, \text{M s}^{-1}
    Show answer and step marks
    1. Correct option: (c) 1.4×10−3 M s−11.4 \times 10^{-3} \, \text{M s}^{-1}, since −Δ[N2O5]Δt=12Δ[NO2]Δt=12×2.8×10−3=1.4×10−3 M s−1-\frac{\Delta[N_2O_5]}{\Delta t} = \frac{1}{2} \frac{\Delta[NO_2]}{\Delta t} = \frac{1}{2} \times 2.8 \times 10^{-3} = 1.4 \times 10^{-3} \, \text{M s}^{-1}.+1
    Answer: (c)
  3. Q3Short answer3 marks
    (a) Differentiate between the order of a reaction and the molecularity of a reaction (any two points). (b) Identify the reaction order if the unit of its rate constant is mol L−1s−1\text{mol L}^{-1} \text{s}^{-1}. (c) Why can molecularity not be zero or fractional?
    Show answer and step marks
    1. (a) First difference, any one of: order is the sum of the powers of concentration terms in the rate law, while molecularity is the number of reacting species colliding simultaneously in an elementary step; order is determined experimentally, while molecularity is theoretical; order can be zero, fractional or negative, while molecularity is always a positive whole number; order applies to elementary and complex reactions, while molecularity applies only to elementary reactions; or any other valid difference.+1
    2. (a) Second difference, a different one from the same list.+1
    3. (b) A rate constant with the unit mol L−1s−1\text{mol L}^{-1} \text{s}^{-1} belongs to a zero order reaction.+0.5
    4. (c) Molecularity is the number of molecules, atoms or ions colliding simultaneously in an elementary step, which can only be a whole number of at least one, never zero or a fraction.+0.5
    Answer: (a) Any two of: order comes from the rate law while molecularity counts colliding species in an elementary step; order is experimental while molecularity is theoretical; order can be zero or fractional while molecularity is a positive integer; molecularity applies only to elementary reactions. (b) Zero order. (c) It counts molecules colliding in an elementary step, which must be a whole number of at least one.

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