Chemical Kinetics Class 12 Notes (2026-27) — CBSE
Class 12 Chemistry Chapter 3 notes: rate of reaction, rate law, order and molecularity, zero and first order kinetics, half-life and the Arrhenius equation.
Chemical Kinetics — Class 12 Chemistry Notes
Chapter Snapshot
Thermodynamics tells you whether a reaction happens; kinetics tells you how fast. This chapter covers the rate of reaction and what affects it, the rate law with order and molecularity, integrated rate equations for zero and first order reactions, half-life, and the temperature dependence of rate through the Arrhenius equation.
Board relevance: ~9 marks with Electrochemistry. Expect a first-order numerical (using k = 2.303/t · log([R]₀/[R])) and an order-vs-molecularity question. Units of k are a favourite one-marker.
Key Concepts & Definitions
Rate of reaction — the change in concentration of a reactant or product per unit time.
- Average rate = Δ[R]/Δt over an interval; instantaneous rate = d[R]/dt at an instant.
- For aA + bB → cC + dD:
rate = −(1/a)d[A]/dt = −(1/b)d[B]/dt = +(1/c)d[C]/dt = +(1/d)d[D]/dt
(the minus signs make reactant rates positive). Unit: mol L⁻¹ s⁻¹.
Factors affecting rate: concentration, temperature, catalyst, surface area, and (for gases) pressure.
Rate law — the experimentally determined dependence of rate on concentration:
rate = k[A]^x[B]^y
Order = x + y (the sum of the powers). It is found experimentally, and may be zero, fractional, or even negative.
Molecularity — the number of species colliding in an elementary step. It is always a whole number (1, 2, or 3), never zero or fractional, and is a theoretical concept.
Order Molecularity
Determined Experimentally From the mechanism
Values Can be 0, fractional, negative Whole number only (1–3)
Applies to Overall reaction Elementary steps only
Rate constant (k) — the rate when all concentrations are unity; independent of concentration but strongly dependent on temperature.
Formulas — Integrated Rate Equations
Zero order (rate independent of concentration)
[R] = [R]₀ − kt ⟹ k = ([R]₀ − [R])/t
- Plot of [R] vs t is a straight line with slope −k.
- Half-life: t½ = [R]₀/2k (proportional to initial concentration).
- Units of k: mol L⁻¹ s⁻¹.
- Examples: decomposition of NH₃ on a hot platinum surface; photochemical reactions.
First order
k = (2.303/t) log([R]₀/[R])
- Plot of log[R] vs t is a straight line with slope −k/2.303.
- Half-life: t½ = 0.693/k — independent of initial concentration (a defining feature).
- Units of k: s⁻¹.
- Examples: radioactive decay; decomposition of N₂O₅.
General units of k for order n: (mol L⁻¹)^{1−n} s⁻¹.
Temperature Dependence
Rate roughly doubles for every 10 °C rise. Quantitatively:
Arrhenius equation:
k = A e^{−Ea/RT}
Taking logarithms:
log k = log A − Ea/(2.303 RT)
- A plot of log k vs 1/T is a straight line of slope −Ea/2.303R — used to find the activation energy.
- For two temperatures:
log(k₂/k₁) = (Ea/2.303R)[(T₂ − T₁)/(T₁T₂)]
Activation energy (Ea) — the minimum extra energy reactants need to form the activated complex (transition state). A lower Ea means a faster reaction.
Collision theory: a reaction occurs when molecules collide with (1) energy ≥ Ea and (2) the correct orientation. Hence rate = PZ·e^{−Ea/RT}, where Z is the collision frequency and P the steric (orientation) factor. Not every collision is effective — only these "fruitful" ones.
Catalyst — provides an alternative path of lower activation energy, increasing the rate. It does not change ΔH or the position of equilibrium; it speeds up forward and backward reactions equally.
Worked Examples
Example 1 — Order and units: A reaction has rate = k[A]²[B]. What is the order and the units of k?
Order = 2 + 1 = 3. Units = (mol L⁻¹)^{1−3} s⁻¹ = mol⁻² L² s⁻¹.
Example 2 — First order: A first order reaction has k = 0.693 min⁻¹. Find its half-life.
t½ = 0.693/k = 0.693/0.693 = 1 minute.
Example 3 — First order concentration: A first order reaction is 50% complete in 20 minutes. How long for 75% completion?
75% complete = two half-lives (50% → 75%), so t = 2 × 20 = 40 minutes.
Example 4 — Using the integrated equation: For a first order reaction with k = 2.303 × 10⁻³ s⁻¹, find the time for the concentration to drop to 1/10 of the initial value.
k = (2.303/t) log(10) → 2.303 × 10⁻³ = (2.303/t)(1) → t = 1000 s.
Example 5 — Activation energy: If the rate doubles when temperature rises from 300 K to 310 K, estimate Ea.
log(2) = (Ea/2.303 × 8.314)[(10)/(300 × 310)] → 0.301 = Ea × 10/(19.147 × 93000) → Ea ≈ 53.6 kJ/mol.
Example 6 — Determining order from data: When [A] is doubled the rate quadruples, and when [B] is doubled the rate is unchanged. Find the rate law and overall order.
Rate ∝ [A]² (doubling gives 2² = 4 times) and rate ∝ [B]⁰ (no effect). So the rate law is rate = k[A]²[B]⁰ = k[A]² and the overall order is 2. This "deduce the order from a table of initial rates" question appears very often — compare experiments in which only one concentration changes at a time.
Important Question Patterns
1. Order/molecularity (2 marks): distinguish them; deduce order from a rate law; give the units of k.
2. First order numerical (3 marks): k = (2.303/t)log([R]₀/[R]); half-life; time for a given percentage completion.
3. Zero order (2–3 marks): [R] = [R]₀ − kt; t½ = [R]₀/2k; sketch the [R] vs t graph.
4. Arrhenius (3 marks): find Ea from two rate constants at two temperatures, or from the slope of a log k vs 1/T plot.
5. Theory (2 marks): collision theory conditions; how a catalyst works and what it does not change.
⚡ Quick Revision
- Rate = −(1/a)d[A]/dt = +(1/c)d[C]/dt (mol L⁻¹ s⁻¹). Affected by concentration, temperature, catalyst, surface area.
- Order (experimental, can be 0/fractional/negative) ≠ molecularity (whole number, elementary steps only).
- Zero order: [R] = [R]₀ − kt; t½ = [R]₀/2k; k in mol L⁻¹ s⁻¹; [R] vs t is linear.
- First order: k = (2.303/t)log([R]₀/[R]); t½ = 0.693/k (independent of [R]₀); k in s⁻¹; log[R] vs t is linear.
- Units of k for order n: (mol L⁻¹)^{1−n} s⁻¹.
- Arrhenius: k = Ae^{−Ea/RT}; log k vs 1/T straight line, slope −Ea/2.303R. Rate ≈ doubles per 10 °C.
- Collision theory: need energy ≥ Ea and correct orientation.
- Catalyst: lowers Ea via an alternative path; does not change ΔH or equilibrium position.
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