CBSE Class 12 Chemistry Chapter 4: Chemical Kinetics NCERT Solutions

Reviewed by NCERT Help Editorial Team · Updated on 29 Jul 2026

NCERT Solutions PDF Class 12 PDF

This chapter delves into the fundamental concepts of Chemical Kinetics for Class 12 Chemistry, as per the CBSE syllabus. The NCERT Solutions provided cover exercises that explain how to determine the order of a reaction and the dimensions of rate constants based on given rate expressions. It also includes problems on calculating initial reaction rates and rates after a certain extent of reaction, using the rate law and the rate constant. These solutions are designed to help students grasp the quantitative aspects of reaction rates and their dependence on reactant concentrations. Mastering these concepts is crucial for understanding reaction mechanisms and predicting reaction behavior under different conditions, aiding in effective exam revision.

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Board	CBSE
Class	Class 12
Subject	Chemiry
Session	2026
Language	English
Type	NCERT Solutions
Chapter	Chapter 4: Chemical Kinetics - NCERT Exercises Solutions

Chapter summary

Chapter 4 of the NCERT Solutions for Class 12 Chemistry focuses on Chemical Kinetics. It provides detailed explanations and solutions for exercises related to determining the order of reactions and the units of rate constants from rate laws. The chapter also covers calculating initial rates and subsequent rates of reaction based on given rate expressions and rate constants, reinforcing the understanding of how concentration affects reaction speed.

Learning outcomes

Determine the order of a chemical reaction from its rate expression.
Calculate the dimensions (units) of the rate constant for different reaction orders.
Apply the rate law to calculate the initial rate of a reaction.
Calculate the rate of a reaction at a specific point when reactant concentrations have changed.
Understand the relationship between rate law, rate constant, and reactant concentrations.

Topics covered

Paper topics

Rate Expression
Order of Reaction
Rate Constant
Dimensions of Rate Constant
Rate Law
Initial Rate Calculation
Rate Calculation after Concentration Change

Important topics

Determining Reaction Order
Calculating Rate Constant Dimensions
Using Rate Law for Rate Calculations
Understanding Concentration Effects on Rate

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Questions and Solutions

Question 4.1

From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants:

(i) $3 NO(g) \rightarrow N_2O(g)$ Rate = $k[NO]^2$

(ii) $H_2O_2(aq) + 3 I^-(aq) + 2 H^+ \rightarrow 2 H_2O(l) + I_3^-$ Rate = $k[H_2O_2][I^-]$

(iii) $CH_3CHO(g) \rightarrow CH_4(g) + CO(g)$ Rate = $k [CH_3CHO]^{3/2}$

(iv) $C_2H_5Cl(g) \rightarrow C_2H_4(g) + HCl(g)$ Rate = $k [C_2H_5Cl]$

Solution:

To determine the order of reaction and the dimensions of the rate constant (k) for each given rate expression:

(i) For the reaction $3 NO(g) \rightarrow N_2O(g)$ , the rate expression is given as Rate = $k[NO]^2$ .

Order of reaction: The order is the sum of the exponents of the concentration terms in the rate law. Here, the exponent of [NO] is 2. Therefore, the order of the reaction is 2.
Dimensions of rate constant (k): The general formula for the dimensions of k is $(\text{concentration})^{1-n} (\text{time})^{-1}$ , where n is the order of the reaction. For n=2, the dimensions are $(\text{mol } L^{-1})^{1-2} (s)^{-1}$ = $(\text{mol } L^{-1})^{-1} (s)^{-1}$ = $L \, mol^{-1} \, s^{-1}$ .

(ii) For the reaction $H_2O_2(aq) + 3 I^-(aq) + 2 H^+ \rightarrow 2 H_2O(l) + I_3^-$ , the rate expression is given as Rate = $k[H_2O_2][I^-]$ .

Order of reaction: The exponent of [H₂O₂] is 1 and the exponent of [I^-] is 1. The total order of the reaction is 1 + 1 = 2.
Dimensions of rate constant (k): For n=2, the dimensions are $(\text{mol } L^{-1})^{1-2} (s)^{-1}$ = $L \, mol^{-1} \, s^{-1}$ .

(iii) For the reaction $CH_3CHO(g) \rightarrow CH_4(g) + CO(g)$ , the rate expression is given as Rate = $k [CH_3CHO]^{3/2}$ .

Order of reaction: The exponent of [CH₃CHO] is 3/2. Therefore, the order of the reaction is 3/2.
Dimensions of rate constant (k): For n=3/2, the dimensions are $(\text{mol } L^{-1})^{1-3/2} (s)^{-1}$ = $(\text{mol } L^{-1})^{-1/2} (s)^{-1}$ = $L^{1/2} \, mol^{-1/2} \, s^{-1}$ .

(iv) For the reaction $C_2H_5Cl(g) \rightarrow C_2H_4(g) + HCl(g)$ , the rate expression is given as Rate = $k [C_2H_5Cl]$ .

Order of reaction: The exponent of [C₂H₅Cl] is 1. Therefore, the order of the reaction is 1.
Dimensions of rate constant (k): For n=1, the dimensions are $(\text{mol } L^{-1})^{1-1} (s)^{-1}$ = $(\text{mol } L^{-1})^{0} (s)^{-1}$ = $s^{-1}$ .

Question 4.2

For the reaction: $2A + B \rightarrow A_2B$ , the rate = $k[A][B]^2$ with $k = 2.0 \times 10^{-6}$ mol^-2 L² s^-1. Calculate the initial rate of the reaction when $[A] = 0.1 \text{ mol } L^{-1}$ , $[B] = 0.2 \text{ mol } L^{-1}$ . Calculate the rate of reaction after $[A]$ is reduced to $0.06 \text{ mol } L^{-1}$ .

Solution:

The given reaction is $2A + B \rightarrow A_2B$ with the rate law Rate = $k[A][B]^2$ and the rate constant $k = 2.0 \times 10^{-6}$ mol^-2 L² s^-1.

1. Calculation of the initial rate of the reaction:

Given initial concentrations are $[A] = 0.1 \text{ mol } L^{-1}$ and $[B] = 0.2 \text{ mol } L^{-1}$ .

Substitute these values into the rate law:

Initial Rate = $k[A][B]^2$

= $(2.0 \times 10^{-6} \text{ mol}^{-2} L^2 s^{-1}) \times (0.1 \text{ mol } L^{-1}) \times (0.2 \text{ mol } L^{-1})^2$

= $(2.0 \times 10^{-6}) \times (0.1) \times (0.04) \text{ mol}^{-2} L^2 s^{-1} \times mol L^{-1} \times mol^2 L^{-2}$

= $8.0 \times 10^{-9} \text{ mol } L^{-1} s^{-1}$

Thus, the initial rate of the reaction is $8.0 \times 10^{-9} \text{ mol } L^{-1} s^{-1}$ .

2. Calculation of the rate of reaction after [A] is reduced to 0.06 mol L^-1:

First, we need to find the concentration of B when [A] has been reduced to $0.06 \text{ mol } L^{-1}$ . According to the stoichiometry of the reaction ( $2A + B \rightarrow A_2B$ ), for every 2 moles of A reacted, 1 mole of B reacts.

Change in [A] = Initial [A] - Final [A] = $0.1 \text{ mol } L^{-1} - 0.06 \text{ mol } L^{-1} = 0.04 \text{ mol } L^{-1}$ .

Amount of B reacted = $\frac{1}{2} \times (\text{Amount of A reacted})$

= $\frac{1}{2} \times 0.04 \text{ mol } L^{-1} = 0.02 \text{ mol } L^{-1}$ .

The concentration of B at this point is:

Final [B] = Initial [B] - Amount of B reacted

= $0.2 \text{ mol } L^{-1} - 0.02 \text{ mol } L^{-1} = 0.18 \text{ mol } L^{-1}$ .

Now, we can calculate the rate of reaction using the rate law with the new concentrations:

Rate = $k[A][B]^2$

= $(2.0 \times 10^{-6} \text{ mol}^{-2} L^2 s^{-1}) \times (0.06 \text{ mol } L^{-1}) \times (0.18 \text{ mol } L^{-1})^2$

= $(2.0 \times 10^{-6}) \times (0.06) \times (0.0324) \text{ mol}^{-2} L^2 s^{-1} \times mol L^{-1} \times mol^2 L^{-2}$

= $3.888 \times 10^{-9} \text{ mol } L^{-1} s^{-1}$

Therefore, the rate of reaction after [A] is reduced to $0.06 \text{ mol } L^{-1}$ is approximately $3.89 \times 10^{-9} \text{ mol } L^{-1} s^{-1}$ .

Common mistakes

Incorrectly calculating the order of reaction from the rate expression.
Errors in determining the units of the rate constant.
Mistakes in substituting values into the rate law for rate calculations.
Not accounting for the change in concentration of reactants when calculating the rate at a later stage.

Revision tips

Practice determining the order and rate constant dimensions for various rate expressions.
Work through numerical problems involving rate calculations, paying close attention to units.
Understand how changes in reactant concentrations affect the reaction rate according to the rate law.
Review the relationship between the rate constant and the rate of reaction.

Practice MCQs

Q1. For a reaction with the rate expression Rat $e = k$ [A]^2[B], what is the overall order of the reaction?

Q2. What are the dimensions of the rate constant for a zero-order reaction?

Q3. If the rate law is Rat $e = k$ [A][B]^2 and $k = 5.0 x 10^{-4} L^2 mol^{-2} s^{-1}$ , what is the order of the reaction with respect to B?

Q4. For the reaction 2A + B -> A2B, if the rate law is Rat $e = k$ [A][B]^2, and [A] = 0.1 M, [B] = 0.2 M, what happens to the rate if [B] is doubled?

Frequently asked questions

What is the order of a reaction?

The order of a reaction is the sum of the exponents of the concentration terms in the rate law expression, indicating how the rate of reaction depends on the concentration of reactants.

How do you find the dimensions of the rate constant?

The dimensions of the rate constant (k) can be determined from the rate law by equating the units of rate (mol L^{-1} s^{-1}) with the units of concentration raised to the power of the reaction order, multiplied by the units of k.

What is the difference between rate and rate constant?

The rate of reaction is the speed at which reactants are consumed or products are formed, expressed in units like mol L^{-1} s^{-1}. The rate constant (k) is a proportionality constant in the rate law that relates the rate of reaction to the concentrations of reactants; its units depend on the order of the reaction.

How can I calculate the initial rate of a reaction?

To calculate the initial rate, substitute the initial concentrations of the reactants into the given rate law expression and multiply by the rate constant.

Why is it important to determine the order of a reaction?

Knowing the order of a reaction is crucial for understanding its mechanism, predicting how changes in concentration will affect the reaction rate, and determining the units of the rate constant.

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