Class 12 Chemistry · NCERT Chapter 3

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Chemical Kinetics Class 12 Chemistry Notes

Chemical kinetics is the study of reaction rates and the factors that control them. This chapter covers rate laws, order and molecularity, integrated rate equations, half-life, the Arrhenius equation, catalysis and collision theory — including the first-order numericals CBSE loves to set.

Class:12Subject:ChemistryCovers:Rate laws · Half-life · ArrheniusExams:CBSE · CUET · NEET

8 Key Formulas2 Solved Examples34 Practice MCQs

DWritten byDeep Narayan

UpdatedOctober 2026

Key Concept Summary

Why is the half-life of a first-order reaction independent of concentration?

For a first-order reaction, t½ = 0.693/k. The time to consume half of the reactant depends only on the rate constant k, because the rate itself is proportional to the remaining concentration — so every successive half-life takes the same time.

Rate of a Reaction

The rate of a reaction is the change in concentration of a reactant or product per unit time. Average rate is measured over a finite interval, while instantaneous rate is the slope of the concentration–time curve at a given instant.

Rate for a reaction aA → bB (negative for reactants, positive for products)

Instantaneous rate = - \frac{1}{a} \frac{d [ R ]}{d t} = \frac{1}{b} \frac{d [ P ]}{d t}

Definition

Rate constant k

The proportionality constant in the rate law. Its units depend on the order of the reaction: for order n they are (mol L⁻¹)¹⁻ⁿ s⁻¹. k does not depend on concentration but does depend on temperature.

Units trap

Zero order: mol L⁻¹ s⁻¹. First order: s⁻¹. Second order: L mol⁻¹ s⁻¹. These appear directly in MCQs.

MCQ Questions

The unit of the rate constant for a first-order reaction is:

The instantaneous rate of reaction is:

Rate Law, Order and Molecularity

Definition

Rate law

The experimentally determined relation Rate = k[A]ˣ[B]ʸ, where the exponents x and y are the orders with respect to A and B. Orders are not related to the stoichiometric coefficients for complex reactions.

Definition

Order of a reaction

The sum of the concentration exponents in the rate law: overall order = x + y. It can be an integer, a fraction or zero, and is found experimentally.

Definition

Molecularity

The number of reactant molecules taking part in an elementary step. It is a small whole number (1, 2, 3) and is defined only for elementary reactions such as the rate-determining step.

Complex reactions proceed in several elementary steps; the slowest one is the rate-determining step.
For an elementary reaction, order equals molecularity; this is not true for complex reactions.
Examples: N₂O₅ → 2NO₂ + ½O₂ is first order overall; decomposition of NH₃ on Pt is zero order.

MCQ Questions

For a reaction with rate = k[A][B]², the overall order is:

Molecularity of a reaction can be:

Integrated Rate Laws: Zero and First Order

Integrated rate equations relate the concentration of a reactant to time. Their linear forms provide graphical tests for the order of a reaction.

Integrated forms for zero- and first-order reactions

Zero order: [A] = [A]_{0} - k t First order: ln [A] = ln [A]_{0} - k t

Zero order: plot [A] vs t gives a straight line with slope −k.
First order: plot ln[A] vs t gives a line with slope −k; intercept ln[A]₀.
First-order examples: decomposition of H₂O₂, N₂O₅, radioactive decay.

Recognition trick

If equal-time segments halve the concentration repeatedly, the reaction is first order. If the concentration drops by a fixed amount per unit time, it is zero order.

MCQ Questions

For a first-order reaction, a plot of ln[A] against t gives:

For a zero-order reaction, the concentration rises or falls:

Half-Life of a Reaction

Definition

Half-life (t½)

The time required for the concentration of a reactant to decrease to half of its initial value. It is a convenient way to compare reaction rates.

Half-life for first- and zero-order reactions

t_{1/2} = \frac{0.693}{k} (1st order), t_{1/2} = \frac{[ A ] _{0}}{2 k} (0th order)

First order: t½ is independent of initial concentration — successive half-lives are equal.
Zero order: t½ proportional to initial concentration — successive half-lives halve.
Radioactive decay uses the first-order law: given t½, k = 0.693/t½.

MCQ favourite

If 2.0 M of a first-order reactant halves to 1.0 M in 10 min, it reaches 0.5 M in a further 10 min — t½ stays the same each time.

MCQ Questions

The half-life of a first-order reaction is 20 min. After 60 min, the fraction remaining is:

For a zero-order reaction, doubling the initial concentration:

Arrhenius Equation and Temperature Dependence

The rate constant rises sharply with temperature. The Arrhenius equation links k to temperature through the activation energy Eₐ — the minimum energy reactant molecules must have to form products.

Arrhenius equation (A = frequency factor)

k = A e^{- E_{a} / R T} or ln k = ln A - \frac{E _{a}}{R T}

Definition

Activation energy Eₐ

The energy barrier between reactants and products. Reactants must acquire at least this energy (usually through collisions) to react. A plot of ln k vs 1/T gives a straight line with slope −Eₐ/R.

Two-temperature form for computing activation energy

lo g \frac{k _{2}}{k _{1}} = \frac{E _{a}}{2.303 R} (\frac{T _{2} - T _{1}}{T _{1} T _{2}})

Rule of thumb

For many reactions, a 10 K rise near room temperature roughly doubles the rate constant — a reflection of the exponential dependence on 1/T.

MCQ Questions

A plot of ln k vs 1/T for a reaction gives:

Q10

Higher activation energy implies:

Collision Theory and Catalysis

Collision theory explains rates: molecules must collide with energy at least Eₐ and with the correct orientation. The rate constant is k = PZ e^(−Eₐ/RT), where Z is the collision frequency and P the steric (orientation) factor, which is < 1 for most reactions.

Only effective collisions — those above the threshold energy and correctly oriented — lead to products.
Increasing temperature raises the fraction of effective collisions far more than the collision frequency.
The activated complex (transition state) is the high-energy arrangement formed momentarily during a collision.

Definition

Catalyst

A substance that increases the rate of a reaction by providing an alternative path of lower activation energy, without itself being consumed and without changing ΔG or the equilibrium constant.

Catalysis in action

Positive catalysts lower Eₐ (e.g. MnO₂ on the decomposition of H₂O₂); inhibitors slow a reaction (e.g. a negative catalyst). Enzymes are biological catalysts with high specificity.

MCQ Questions

Q11

A catalyst increases the rate of a reaction by:

Q12

In collision theory, the factor P accounts for:

NCERT Exercise Questions

From the textbook

NCERT questions for this chapter, in textbook order. Answers are given at a glance below each set.

Exercises (NCERT Q1–Q6)

1
For the reaction R → P, the concentration of a reactant changes from 0.03 M to 0.02 M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.
2
In a reaction, 2A → Products, the concentration of A decreases from 0.5 mol L⁻¹ to 0.4 mol L⁻¹ in 10 minutes. Calculate the rate during this interval.
3
The decomposition of NH3 on platinum surface is zero order reaction. What are the rates of production of N2 and H2 if k = 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹?
4
The decomposition of dimethyl ether leads to the formation of CH4, H2 and CO and the reaction rate is given by Rate = k [CH3OCH3]^(3/2). If the pressure is measured in bar and time in minutes, what are the units of rate and rate constant?

Answers at a glance

1
Average rate = 4.0 × 10⁻⁴ M min⁻¹ = 6.7 × 10⁻⁶ M s⁻¹
2
Rate = 5.0 × 10⁻³ mol L⁻¹ min⁻¹
3
Rate of N2 formation = 1.25 × 10⁻⁴ mol L⁻¹ s⁻¹; Rate of H2 formation = 3.75 × 10⁻⁴ mol L⁻¹ s⁻¹
4
Unit of rate = bar min⁻¹; Unit of rate constant = bar^(-1/2) min⁻¹

CBSE Pattern Questions

How CBSE asks this chapter

CBSE typically asks 2-3 questions from Chemical Kinetics, focusing on numerical problems of rate law, order of reaction, integrated rate equations (zero and first order), half‑life and Arrhenius equation.

MCQ Questions

Q13

For the reaction R → P, the concentration of a reactant changes from 0.03 M to 0.02 M in 25 minutes. Calculate the average rate of reaction.

Q14

The decomposition of NH3 on platinum surface is zero order reaction. What is the rate of production of NH3 when k = 2.5 × 10−4 mol L−1 s−1?

Q15

The rate constant for a first order reaction is 60 s−1. How much time will it take to reduce the initial concentration to 1/16th of its initial value?

Q16

The activation energy for a reaction is zero. The rate constant of the reaction:

1 Mark Questions

1
Define rate of a reaction.
2
What is the order of a reaction?
3
Write the integrated rate equation for a zero order reaction.

2 Mark Questions

1
For the reaction 2A + B → A2B, the rate = k[A][B]2 with k = 2.0 × 10−6 M−2 s−1. Calculate the initial rate when [A] = 0.1 M, [B] = 0.2 M.
2
Show that in a first order reaction, time required for completion of 99.9% is 10 times the half‑life.

3 Mark Questions

1
Derive the integrated rate equation for a first order reaction.
2
Explain the effect of temperature on the rate constant with the help of Arrhenius equation.

5 Mark Questions

1
The decomposition of dimethyl ether leads to the formation of CH4, H2 and CO and the reaction rate is given by Rate = k [CH3OCH3]3/2. If the rate is monitored by increase in pressure in a closed vessel, derive the expression for the rate in terms of total pressure of the system.
2
For the first order thermal decomposition of N2O5(g) at constant volume, 2N2O5(g) → 4NO2(g) + O2(g), calculate the rate constant when the initial pressure of N2O5 is 0.2 atm and after 50 s it becomes 0.182 atm.

PDF: Assignments & Tests

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Download topic-wise assignments and full CBSE-pattern tests with answer keys for this chapter.

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Class 12 Chemistry — downloadable assignments and CBSE-pattern tests (PDF).

Assignment 1 — Topic-wise Practice Set A

Topic-wise practice covering the first half of the chapter, with answer key.

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Assignment 2 — Topic-wise Practice Set B

Topic-wise practice covering the second half of the chapter, with answer key.

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CBSE Pattern Test 1

Full CBSE-pattern test (90 minutes, 40 marks) with answer key.

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CBSE Pattern Test 2

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Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Rate of reaction

Rate = - \frac{1}{a} \frac{d [ A ]}{d t} = \frac{1}{b} \frac{d [ B ]}{d t}

Instantaneous rate is a derivative; units mol L⁻¹ s⁻¹.

Rate law

Rate = k [A]^{x} [B]^{y}

x, y are determined experimentally.

Zero-order integrated law

[A] = [A]_{0} - k t

[A] falls linearly; unit of k is mol L⁻¹ s⁻¹.

First-order integrated law

ln [A] = ln [A]_{0} - k t

∘ Graphical form: ln[A] vs t is a straight line.

First-order half-life

t_{1/2} = \frac{0.693}{k}

Independent of initial concentration.

Zero-order half-life

t_{1/2} = \frac{[ A ] _{0}}{2 k}

Proportional to initial concentration.

Arrhenius equation

k = A e^{- E_{a} / R T}

A = frequency factor, Eₐ = activation energy.

Two-temperature form

lo g \frac{k _{2}}{k _{1}} = \frac{E _{a}}{2.303 R} \frac{T _{2} - T _{1}}{T _{1} T _{2}}

Used to find Eₐ from k at two temperatures.

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

Order vs molecularity — definitional difference is a fixture of 2-mark and MCQ sections.
Units of the rate constant k for reactions of order n: (mol L⁻¹)¹⁻ⁿ s⁻¹.
Half-life of first-order reactions (t½ = 0.693/k) and independence from initial concentration.
Graphical identification: ln[A] vs t straight line for first order, [A] vs t for zero order.
Using the two-temperature Arrhenius form to compute activation energy (3 marks).
Effect of a catalyst: lower activation energy, unchanged equilibrium constant.
Collision theory: effective collisions, threshold energy, orientation factor and P·Z in k = PZAe^(−Ea/RT).
Elementary vs complex reactions and rate-determining step.

Practice Numericals

Step-by-step worked examples

Standard exam-level problems solved with clear conceptual breakdowns.

Half-life of a first-order reaction

Problem StatementA first-order reaction has a rate constant of 0.0231 min⁻¹. Calculate its half-life and the time to reach 90% completion.

Step-by-Step Solution

1t½ = 0.693 / k = 0.693 / 0.0231 ≈ 30 min.
2For 90% completion, [A] = 0.1[A]₀, so ln(0.1) = −kt → t = 2.303 log₁₀(10) / k.
3t = 2.303 × 1 / 0.0231 ≈ 99.7 min.

Final Answer

t½ ≈ 30 min; t(90%) ≈ 100 min

Activation energy from two rate constants

Problem Statementk₁ = 1.5 × 10⁻⁴ s⁻¹ at 600 K and k₂ = 6.0 × 10⁻⁴ s⁻¹ at 650 K. Find Eₐ (R = 8.314 J mol⁻¹ K⁻¹).

Step-by-Step Solution

1log₁₀(k₂/k₁) = (Eₐ/2.303R) × (T₂ − T₁)/(T₁T₂).
2log₁₀(4.0) = (Eₐ / (2.303 × 8.314)) × (50)/(600 × 650).
3Eₐ = 0.6021 × 2.303 × 8.314 × 600 × 650 / 50.

Final Answer

Eₐ ≈ 89.9 kJ mol⁻¹

FAQ

Frequently asked questions

What is the difference between order and molecularity?

Order is the sum of concentration exponents in the experimentally determined rate law and can be fractional or zero. Molecularity is the number of reacting molecules in an elementary step, is always a small whole number, and applies only to elementary reactions.

Why does the rate constant nearly double for a 10 K rise near room temperature?

From the Arrhenius equation, k ∝ e^(−Ea/RT). A 10 K rise increases the fraction of molecules crossing the activation-energy barrier steeply — roughly a two-fold increase for typical activation energies — even though the collision frequency itself hardly changes.

Does a catalyst change the equilibrium of a reaction?

No. A catalyst speeds up both forward and backward reactions equally by lowering the activation energy on an alternative path, so the equilibrium constant K and the position of equilibrium are unchanged.

Can the order of a reaction be zero?

Yes. In zero-order reactions (e.g. decomposition of NH₃ on a platinum surface, or enzyme-saturated reactions) the rate is independent of the reactant concentration because a surface or catalyst active site is saturated. The half-life then equals [A]₀/2k, which grows with initial concentration.

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