CBSE Class 12 Chemistry Chapter 8: Chemical Kinetics NCERT Solutions

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

NCERT Solutions PDF Class 12 PDF

This chapter, Chemical Kinetics, delves into the fundamental principles governing the rates of chemical reactions. The NCERT Solutions for Class 12 Chemistry Chapter 8 provide a detailed exploration of concepts such as reaction rates, factors affecting them, and the determination of the order of a reaction. Students will learn to interpret rate expressions, calculate the dimensions of rate constants, and apply these principles to solve numerical problems. The solutions cover various types of reactions, including those with different rate laws and orders. Understanding these concepts is crucial for predicting reaction speeds and designing chemical processes. These solutions are designed to aid students in mastering the chapter's content, reinforcing their understanding through step-by-step problem-solving, and preparing effectively for their board examinations.

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Board	CBSE
Class	Class 12
Subject	Chemistry
Session	2026
Language	English
Type	NCERT Solutions
Chapter	Chapter 8

Chapter summary

Chapter 8 of NCERT Class 12 Chemistry, Chemical Kinetics, focuses on the study of reaction rates and mechanisms. The provided NCERT Solutions offer clear explanations and step-by-step solutions for exercises related to determining the order of reactions, calculating rate constants, and understanding their dimensions. It covers various rate laws and their implications, helping students grasp the quantitative aspects of chemical reactions.

Learning outcomes

Determine the order of a chemical reaction from its rate expression.
Calculate the dimensions of rate constants for different reaction orders.
Interpret rate laws and apply them to predict reaction rates.
Solve numerical problems involving initial rates and changes in reactant concentrations.
Understand the relationship between rate constant, concentration, and time.

Topics covered

Paper topics

Rate of Reaction
Rate Expression
Order of Reaction
Rate Constant
Dimensions of Rate Constant
Rate Law
First Order Reaction
Second Order Reaction
Third Order Reaction
Fractional Order Reaction
Initial Rate Calculation
Rate Change with Concentration

Important topics

Determining Order of Reaction
Calculating Rate Constant Dimensions
Applying Rate Laws to Calculate Rates
Understanding Rate Expressions
Initial Rate Calculations

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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) $3 NO(g) → N_2O(g) Rate = k[NO]^2$

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

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

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

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 $Rate = k[NO]^2$ .

The order of the reaction 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.

To find the dimensions of k, we rearrange the rate expression: $k = \frac{Rate}{[NO]^2}$

The units for rate are typically $mol L^{-1} s^{-1}$ , and the units for concentration are $mol L^{-1}$ .

So, the dimensions of k are: $\frac{mol L^{-1} s^{-1}}{(mol L^{-1})^2} = \frac{mol L^{-1} s^{-1}}{mol^2 L^{-2}} = L \ mol^{-1} s^{-1}$ .

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

The order of the reaction is the sum of the exponents of [H2O2] (which is 1) and [I-] (which is 1). Therefore, the order of the reaction is 1 + 1 = 2.

To find the dimensions of k, we rearrange the rate expression: $k = \frac{Rate}{[H_2O_2][I^-]}$

The dimensions of k are: $\frac{mol L^{-1} s^{-1}}{(mol L^{-1})(mol L^{-1})} = \frac{mol L^{-1} s^{-1}}{mol^2 L^{-2}} = L \ mol^{-1} s^{-1}$ .

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

The order of the reaction is the exponent of the concentration term [CH3CHO], which is 3/2. Therefore, the order of the reaction is 3/2.

To find the dimensions of k, we rearrange the rate expression: $k = \frac{Rate}{[CH_3CHO]^{3/2}}$

The dimensions of k are: $\frac{mol L^{-1} s^{-1}}{(mol L^{-1})^{3/2}} = \frac{mol L^{-1} s^{-1}}{mol^{3/2} L^{-3/2}} = L^{1/2} \ mol^{-1/2} s^{-1}$ .

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

The order of the reaction is the exponent of the concentration term [C2H5Cl], which is 1. Therefore, the order of the reaction is 1.

To find the dimensions of k, we rearrange the rate expression: $k = \frac{Rate}{[C_2H_5Cl]}$

The dimensions of k are: $\frac{mol L^{-1} s^{-1}}{mol L^{-1}} = s^{-1}$ .

Question 4.2

For the reaction:

$2A\,+\,B o A_2B$

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

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:

$Initial Rate = k[A][B]^2 = (2.0 \times 10^{-6} \text{ L}^2 \text{ mol}^{-2} \text{ s}^{-1}) \times (0.1 \text{ mol } L^{-1}) \times (0.2 \text{ mol } L^{-1})^2$

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

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

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

First, we need to determine the concentration of B when [A] has been reduced to $0.06 \text{ mol } L^{-1}$ . According to the stoichiometry of the reaction, 2 moles of A react with 1 mole of B.

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

The amount of B reacted is half the amount of A reacted:

Amount of B reacted = $\frac{1}{2} \times (\text{Change in } [A]) = \frac{1}{2} \times 0.04 \text{ mol } L^{-1} = 0.02 \text{ mol } L^{-1}$ .

The concentration of B remaining 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 the reaction with the new concentrations:

$Rate = k[A][B]^2 = (2.0 \times 10^{-6} \text{ L}^2 \text{ mol}^{-2} \text{ s}^{-1}) \times (0.06 \text{ mol } L^{-1}) \times (0.18 \text{ mol } L^{-1})^2$

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

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

Therefore, the initial rate of the reaction is $8.0 \times 10^{-9} \text{ mol } L^{-1} \text{ s}^{-1}$ , and the rate of the reaction when [A] is reduced to $0.06 \text{ mol } L^{-1}$ is approximately $3.89 \times 10^{-9} \text{ mol } L^{-1} \text{ s}^{-1}$ .

Common mistakes

Incorrectly calculating the order of reaction from the rate law.
Errors in unit conversions or dimensional analysis for the rate constant.
Misinterpreting the rate law when exponents are fractional.
Calculation errors when determining the rate after a change in concentration.

Revision tips

Practice identifying the order of reaction directly from the given rate expression.
Pay close attention to the units of the rate constant to determine its dimensions.
Work through all numerical problems, especially those involving initial rates and subsequent concentration changes.
Review the relationship between rate law, rate constant, and reactant concentrations before attempting problems.

Practice MCQs

Q1. For a reaction with the rate expression Rat $e = k$ [A]^x[B]^y, the order of the reaction is:

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

Q3. If the rate of a reaction is given by Rat $e = k$ [CH3CHO]^(3/2), what is the order of the reaction?

Q4. For the reaction 2A + B -> A2B, if the rate law is Rat $e = k$ [A][B]^2, and [A] = 0.1 mol L^-1, [B] = 0.2 mol L^-1, what is the initial rate if $k = 2.0 x 10^-6 L^2 mol^-2 s^-1$ ?

Frequently asked questions

What is the main focus of Chapter 8, Chemical Kinetics, in Class 12 Chemistry?

Chapter 8 focuses on the study of the rates of chemical reactions, the factors that influence these rates, and the mechanisms by which reactions occur.

How do the NCERT Solutions help in understanding the order of a reaction?

The solutions provide step-by-step methods to determine the order of a reaction by analyzing its rate expression, including cases with fractional orders.

What are the dimensions of the rate constant, and how are they calculated?

The dimensions of the rate constant vary depending on the order of the reaction. The solutions show how to derive these dimensions using the units of rate and concentration.

Can these solutions help in solving numerical problems related to reaction rates?

Yes, the solutions include numerical problems that demonstrate how to calculate initial rates and predict reaction rates after changes in reactant concentrations.

Are the mathematical expressions in the questions preserved in the solutions?

Yes, all mathematical expressions, symbols, and equations from the original questions are preserved exactly in the rewritten solutions.

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