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Chemical Kinetics – Class 12 Chemistry notes, important questions and MCQs

NCERT Class 12 Chemistry, chapter 3. Notes, key terms, important questions with answers and MCQs.

Chapter overview

Chemical kinetics studies the rate of reactions, the factors that affect it, and how rate depends on concentration (rate law and order).

Rate and order

Rate = −d[R]/dt = d[P]/dt; rate law: rate = k[A]ˣ[B]ʸ; order = x + y (found experimentally).
Zero order: [R] = [R]₀ − kt; t½ = [R]₀/(2k).
First order: k = (2.303/t) log([R]₀/[R]); t½ = 0.693/k (independent of concentration).

Temperature and catalysts

Arrhenius equation: k = A e^(−Ea/RT); rate roughly doubles for a 10 °C rise.
log(k₂/k₁) = (Ea/2.303R)(1/T₁ − 1/T₂).
A catalyst lowers activation energy and is not consumed; it does not change ΔG or equilibrium.

Key terms

Activation energy: minimum extra energy reactants need to form products
Half-life: time for the concentration to fall to half
Order of reaction: sum of the powers of concentration terms in the experimental rate law

Formulas

t½ = 0.693/k (first order)
k = A e^(−Ea/RT)
k = (2.303/t) log([R]₀/[R])

Important questions with answers

How does a catalyst increase the rate of a reaction?2 marks

It provides an alternative pathway with lower activation energy, so more molecules have enough energy to react. It does not change ΔG or the equilibrium constant.

Show that the half-life of a first-order reaction is independent of initial concentration.3 marks

Integrated rate law: k = (2.303/t) log([R]₀/[R]). At t = t½, [R] = [R]₀/2, so k = (2.303/t½) log 2, giving t½ = 0.693/k, which does not contain [R]₀.

Write the Arrhenius equation and explain the effect of temperature on rate. The rate constant of a reaction doubles when temperature rises from 300 K to 310 K. Calculate the activation energy (R = 8.314 J/mol K).5 marks

k = A e^(−Ea/RT). Raising temperature increases the fraction of molecules with energy ≥ Ea (area under the Maxwell–Boltzmann curve beyond Ea), so the rate rises sharply; typically it about doubles for a 10 K rise. log(k₂/k₁) = (Ea/2.303R)(T₂ − T₁)/(T₁T₂). log 2 = 0.301 = (Ea/(2.303 × 8.314)) × (10/93,000). Ea = 0.301 × 2.303 × 8.314 × 9,300 ≈ 53,600 J/mol ≈ 53.6 kJ/mol.

Quick revision

Order is experimental.
First-order half-life is constant.
Catalyst lowers Ea.

Exam tips

Units of k tell the order: mol L⁻¹ s⁻¹ (zero), s⁻¹ (first), L mol⁻¹ s⁻¹ (second).
Order is found by experiment; molecularity applies only to elementary steps and is always a whole number (1, 2 or 3).

Practice MCQs

A catalyst
1. increases Ea
2. lowers Ea
3. changes ΔG
4. is used up
Show answer
Answer: (b) lowers Ea. Lowers activation energy.
Unit of k for a first-order reaction is
1. mol L⁻¹ s⁻¹
2. mol s
3. L mol⁻¹ s⁻¹
4. s⁻¹
Show answer
Answer: (d) s⁻¹. s⁻¹.
Half-life of a first-order reaction with k = 0.0693 min⁻¹ is
1. 1 min
2. 0.1 min
3. 100 min
4. 10 min
Show answer
Answer: (d) 10 min. 0.693/0.0693.
For rate = k[A][B]², the overall order is
1. 1
2. 2
3. 3
4. 4
Show answer
Answer: (c) 3. 1 + 2.
Unit of rate constant for a second-order reaction is
1. s⁻¹
2. L mol⁻¹ s⁻¹
3. mol L⁻¹ s⁻¹
4. L² mol⁻² s⁻¹
Show answer
Answer: (b) L mol⁻¹ s⁻¹. (mol/L)^(1−n) s⁻¹.
A first-order reaction is 75% complete in 60 min. Its half-life is
1. 15 min
2. 30 min
3. 45 min
4. 60 min
Show answer
Answer: (b) 30 min. 75% = two half-lives.
Which is NOT changed by a catalyst?
1. rate
2. activation energy
3. equilibrium constant
4. reaction path
Show answer
Answer: (c) equilibrium constant. K depends on ΔG only.
A first-order reaction is 75% complete in 60 min. Its half-life is
1. 15 min
2. 30 min
3. 45 min
4. 60 min
Show answer
Answer: (b) 30 min. 75% = two half-lives.

All chapters of Class 12 Chemistry

Written by Kundali IQ from standard, well-established facts to help with revision. This is not an official NCERT publication; always read your textbook too.