NCERT Books for Class 12 Chemistry Chapter 4 - Chemical Kinetics

In the Class 12 board exam, you can expect numerical problems based on:

• Calculating the half-life (t₁/₂) of zero and first-order reactions.
• Using the integrated rate equation for a first-order reaction to find the time required for a certain percentage of completion (e.g., 99.9%).
• Determining the order of a reaction and the rate constant (k) from a given set of experimental concentration and rate data.
• Applying the Arrhenius equation to calculate activation energy or the rate constant at a different temperature.

These questions are often asked for 3 marks.

Yes, two derivations from this chapter are very important for the CBSE board exam. Students must practice deriving the:

• Integrated rate equation for a zero-order reaction (k = ([R]₀ - [R])/t).
• Integrated rate equation for a first-order reaction (k = (2.303/t) log([R]₀/[R])).

These derivations are frequently asked as 2 or 3-mark questions.

This is a critical concept for scoring well. Remember these key differences:

• Basis: Order is an experimental quantity determined from the rate law. Molecularity is a theoretical concept for elementary reactions only.
• Values: Order can be a whole number, zero, or even a fraction. Molecularity must be a positive integer (1, 2, or 3).
• Applicability: Order applies to both elementary and complex reactions. Molecularity is only defined for elementary (single-step) reactions.

Answering with these distinctions helps secure full marks in theory-based questions.

The half-life of a first-order reaction is independent of the initial reactant concentration because its formula, t₁/₂ = 0.693/k, does not contain any concentration term ([R]₀). This is a crucial concept for HOTS (Higher Order Thinking Skills) questions. For example, if a problem states that the initial concentration of a first-order reactant is doubled, its half-life remains unchanged. Recognising this can help you solve MCQs and short-answer questions quickly and accurately.

A catalyst increases the rate of a reaction by providing an alternative reaction pathway with a lower activation energy (Ea). It does not get consumed in the reaction and does not alter the equilibrium constant (K) or Gibbs free energy (ΔG). In board exams, this is often tested by asking you to:

• Interpret an energy profile diagram, showing the lowered energy barrier in the presence of a catalyst.
• Explain why a catalyst increases the rate of both forward and reverse reactions equally.
• Answer conceptual questions on the characteristics of a catalyst.

The most effective strategy is to look for key information in the problem. First, check the units of the rate constant (k) if provided:

• For zero-order, the unit is mol L⁻¹ s⁻¹.
• For first-order, the unit is s⁻¹.
• For second-order, the unit is L mol⁻¹ s⁻¹.

If k is not given, analyse the experimental data table. See how the rate changes when the concentration of a reactant is doubled or halved while keeping others constant. This will help you determine the power to which each reactant concentration is raised in the rate law, thus finding the overall order.

Graph-based questions are a common feature. Be prepared to identify the order of a reaction or calculate values from plots such as:

• Zero-Order: A plot of [R] vs. time gives a straight line with a slope of -k.
• First-Order: A plot of ln[R] vs. time gives a straight line with a slope of -k, or a plot of log[R] vs. time gives a slope of -k/2.303.
• Arrhenius Equation: A plot of ln(k) vs. 1/T gives a straight line with a slope of -Ea/R, from which activation energy (Ea) can be calculated.