Class 12 Chemistry Chapter 3 Chemical Kinetics – Extra Questions with Answers

These extra practice questions for Class 12 Chemistry Chapter 3 – Chemical Kinetics go beyond the NCERT textbook exercises to reinforce rate laws and order of reaction, integrated rate equations, half-life, and the Arrhenius equation. Useful for board exam revision and quick concept checks.

Last Updated: September 23, 2026

Very Short Answer Type Questions (1 Mark)

Q1. Define the rate of a chemical reaction.
Ans: The rate of a reaction is the change in concentration of any one reactant or product per unit time, at a given instant — it is always expressed as a positive quantity, with reactant concentration terms carrying a negative sign since they decrease over time.

Q2. What is meant by the order of a reaction?
Ans: The order of a reaction is the sum of the powers to which the concentration terms are raised in the experimentally determined rate law. It can be zero, a whole number, or a fraction, and must be determined experimentally — it cannot be deduced from the balanced chemical equation alone.

Q3. What is the half-life of a first-order reaction?
Ans: For a first-order reaction, the half-life (t_1/2=0.693/k) is independent of the initial concentration — it takes the same amount of time to reduce the concentration from any starting value to half that value.

Q4. What is activation energy?
Ans: Activation energy (E_a) is the minimum extra energy that reactant molecules must possess, over their average energy, for a collision between them to result in a successful reaction (i.e. to form the activated complex/transition state).

Short Answer Type Questions (2–3 Marks)

Q5. Distinguish between the order of a reaction and the molecularity of a reaction.
Ans: Order is an experimentally determined quantity obtained from the rate law — it can be zero, fractional, or a whole number, and applies to the overall reaction (or to individual steps of a complex mechanism). Molecularity is a theoretical concept — the number of reacting species (atoms, ions or molecules) that must collide simultaneously to bring about a single elementary reaction step; it is always a whole number (1, 2, or rarely 3) and is meaningful only for a single elementary step, not for a complex multi-step reaction as a whole.

Q6. Explain why a reaction with a small activation energy is generally fast, while one with a large activation energy is generally slow.
Ans: According to the Boltzmann distribution of molecular energies, only a fraction of molecules possess energy equal to or greater than the activation energy E_a at any given temperature. If E_a is small, a large fraction of the colliding molecules already have enough energy to react, so the reaction proceeds quickly. If E_a is large, only a very small fraction of collisions have enough energy to be effective, so the reaction is slow — this fraction is captured mathematically by the exponential term e^−E_a/RT in the Arrhenius equation.

Q7. For a zero-order reaction, show that the rate constant has the same units as the rate of reaction, and describe how concentration varies with time.
Ans: For a zero-order reaction, rate=k[A]⁰=k, so k itself has units of concentration/time (e.g. mol L⁻¹s⁻¹), identical to the units of rate, since the concentration term carries no power. Integrating −d[A]/dt=k gives [A]_t=[A]₀−kt — concentration decreases linearly with time for a zero-order reaction.

Higher Order Thinking Skills (HOTS)

Q8. A first-order reaction has a rate constant of 1.15×10⁻³s⁻¹. How long will it take for 5g of this reactant to reduce to 3g?
Ans: Since t=(1/k)ln([A]₀/[A]_t) and the ratio [A]₀/[A]_t can be taken directly as a mass ratio for a first-order reaction (concentration is proportional to mass at constant volume): t=(1/1.15×10⁻³)×ln(5/3)≈(1/1.15×10⁻³)×0.5108≈444s (≈7.4 minutes).

Q9. Two students run the same reaction at the same temperature but disagree about its activation energy after fitting their own rate-constant-vs-temperature data to the Arrhenius equation. Explain what feature of their data (plotted as ln k vs 1/T) would let them resolve the disagreement without repeating the experiment, and what a steeper slope would mean physically.
Ans: Since ln k=ln A−(E_a/R)(1/T), plotting ln k against 1/T should give a straight line whose slope equals −E_a/R; both students can plot their own (k,T) data pairs this way and read off E_a directly from the slope (E_a=−slope×R), rather than relying on a single two-point calculation that is more sensitive to experimental error in any one measurement. A steeper (more negative) slope corresponds to a larger activation energy — physically, it means the reaction’s rate constant is more sensitive to temperature change, since a higher energy barrier means a proportionally larger fraction of molecules gain (or lose) the ability to react for a given change in T.

Written by Satish

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