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Tamil Nadu Class 12 Chemistry — Chemical Kinetics Notes & Important Questions

Samacheer Kalvi Class 12 Chemistry Chemical Kinetics — rate of reaction, rate law, integrated rate equations, Arrhenius equation, activation energy, collision theory.

Updated 18 May 202611 min read

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Quick Summary

Chemical kinetics studies how fast a reaction occurs and the factors that control that rate. In TN HSE board exams, the most tested areas are: rate law and order determination from data tables, first-order integrated equations, half-life calculations, Arrhenius equation numericals, and graph-based questions. This guide covers all of those systematically.


1. Core Definitions

TermDefinition
Rate of reactionChange in concentration of reactant or product per unit time; rate = −d[A]/dt = +d[P]/dt
Instantaneous rateRate at a specific moment; slope of tangent on concentration–time graph
Average rateChange over a time interval; Δ[A]/Δt
Rate constant, kProportionality constant in rate law; depends on T and Ea, not concentration
OrderSum of powers of concentration terms in the experimentally derived rate law
MolecularityNumber of reacting molecules in an elementary (single-step) reaction
Activation energy, EaMinimum energy that colliding molecules must have to react

2. Rate Law and Reaction Order

For a general reaction:

aA + bB → products

Experimental rate law:

Rate = k[A]^x [B]^y
  • x = order with respect to A
  • y = order with respect to B
  • Overall order = x + y

Critical rule: x and y are determined experimentally — never assume x = a or y = b from stoichiometry unless the reaction is elementary.

Determining Order from Data

Method — compare experiments where one concentration changes and the other is held constant.

Example data:

Experiment[A] / mol L⁻¹[B] / mol L⁻¹Rate / mol L⁻¹ s⁻¹
10.100.102.0 × 10⁻³
20.200.104.0 × 10⁻³
30.100.202.0 × 10⁻³

From Exp 1 and 2 ([B] constant, [A] doubles): rate doubles → x = 1 (first order in A) From Exp 1 and 3 ([A] constant, [B] doubles): rate unchanged → y = 0 (zero order in B) Overall order = 1. Rate = k[A].


3. Integrated Rate Equations

Zero-Order Reaction

Rate = k (constant, independent of concentration)

[A]t = [A]₀ − kt
  • Plot of [A] vs t is a straight line with slope = −k
  • Half-life: t½ = [A]₀ / 2k (depends on initial concentration)
  • Units of k: mol L⁻¹ s⁻¹

Example: Decomposition of N₂O on hot platinum surface.

First-Order Reaction

Rate = k[A]

Integration gives:

ln[A]t = ln[A]₀ − kt or k = (2.303/t) log([A]₀/[A]t)
  • Plot of log[A] vs t is a straight line with slope = −k/2.303
  • Half-life: t½ = 0.693/k (independent of [A]₀ — this is the hallmark of first order)
  • Units of k: s⁻¹ (or min⁻¹, etc.)

Examples: radioactive decay, decomposition of SO₂Cl₂, first-order hydrolysis of esters.

Pseudo First-Order Reactions

When a higher-order reaction has one reactant in large excess (effectively constant concentration), it behaves as first order. The apparent rate constant k' = k[B]_excess is called the pseudo first-order rate constant.

Classic example: Hydrolysis of methyl acetate in dilute HCl — water is in vast excess, so rate depends only on ester concentration.


4. Arrhenius Equation

k = A · e^(−Ea/RT)

In logarithmic form:

log k = log A − Ea / (2.303 RT)
  • A is the frequency factor (pre-exponential factor) — accounts for collision frequency and orientation
  • Ea is activation energy in J mol⁻¹ (or kJ mol⁻¹)
  • R = 8.314 J mol⁻¹ K⁻¹
  • T is temperature in Kelvin

Finding Ea from Two Temperatures

log(k₂/k₁) = (Ea / 2.303R) × (T₂ − T₁) / (T₁T₂)

If k doubles for a 10°C rise (T₁ = 300 K, T₂ = 310 K):

log(2) = (Ea / 2.303 × 8.314) × (10 / 93000) 0.3010 = Ea / 19.14 × 1.075 × 10⁻⁴ Ea ≈ 53,600 J mol⁻¹ ≈ 53.6 kJ mol⁻¹

5. Collision Theory

For a reaction to occur, molecules must: 1. Collide — sufficient collision frequency 2. Collide with energy ≥ Ea — threshold energy 3. Collide with proper orientation (steric factor p)

k = p · Z · e^(−Ea/RT)

where Z is the collision frequency.

Energy distribution (Maxwell-Boltzmann curve):

  • Area under the curve beyond Ea represents fraction of effective molecules
  • Higher T → peak shifts right → larger fraction exceeds Ea → faster rate

6. Effect of Catalyst

A catalyst provides an alternative reaction pathway with lower activation energy.

  • It does NOT change ΔH, ΔG, or the equilibrium constant K
  • It increases both forward and reverse rates equally
  • It reaches the equilibrium faster — does not shift equilibrium position
  • The activated complex is different (lower energy) on the catalysed path

Mechanism of heterogeneous catalysis: 1. Adsorption of reactants on catalyst surface 2. Formation of surface intermediate 3. Reaction between adsorbed species 4. Desorption of products from surface


7. Worked Numericals

Problem 1 — Determining Rate Law from Experimental Data

Data:

Exp[NO] / M[Cl₂] / MInitial rate / M s⁻¹
10.100.101.2 × 10⁻³
20.100.202.4 × 10⁻³
30.200.104.8 × 10⁻³

Find order w.r.t. each reactant and k.

Step 1 — Order w.r.t. Cl₂ (Exp 1 vs 2, [NO] constant): Rate₂/Rate₁ = (2.4/1.2) = 2 = (0.20/0.10)^y = 2^y → y = 1 Step 2 — Order w.r.t. NO (Exp 1 vs 3, [Cl₂] constant): Rate₃/Rate₁ = (4.8/1.2) = 4 = (0.20/0.10)^x = 2^x → x = 2 Step 3 — Overall order = 2 + 1 = 3 Rate = k[NO]²[Cl₂] Step 4 — Find k from Exp 1: 1.2 × 10⁻³ = k × (0.10)² × 0.10 k = 1.2 × 10⁻³ / 1.0 × 10⁻³ = 1.2 M⁻² s⁻¹

Problem 2 — First-Order Half-Life and Time for 75% Completion

GivenA first-order reaction has k = 1.386 × 10⁻² min⁻¹

Find: (a) t½, (b) time for 75% completion

(a) t½ = 0.693 / k = 0.693 / (1.386 × 10⁻²) = 50 min (b) 75% completion means 25% remains: [A]t/[A]₀ = 25/100 = 0.25 = (1/2)² = two half-lives Time = 2 × t½ = 2 × 50 = 100 min

Alternatively, using the formula:

t = (2.303/k) log([A]₀/[A]t) = (2.303 / 0.01386) log(100/25) = 166.2 × log 4 = 166.2 × 0.6021 = 100.1 min ≈ 100 min ✓

Problem 3 — Arrhenius Equation (Effect of Temperature)

GivenEa = 40 kJ mol⁻¹ = 40,000 J mol⁻¹, T₁ = 300 K, T₂ = 310 K

Find: Ratio k₂/k₁

log(k₂/k₁) = Ea/2.303R × (T₂ − T₁)/(T₁T₂) = 40000 / (2.303 × 8.314) × 10 / (300 × 310) = 40000 / 19.14 × 10 / 93000 = 2089.86 × 1.075 × 10⁻⁴ = 0.2247 k₂/k₁ = 10^0.2247 = 1.677 ≈ 1.68

The rate increases by a factor of 1.68 for a 10°C rise when Ea = 40 kJ mol⁻¹.


Problem 4 — Finding Ea from Rate Constants at Two Temperatures

Givenk₁ = 2.0 × 10⁻³ s⁻¹ at 300 K, k₂ = 8.0 × 10⁻³ s⁻¹ at 320 K

Find: Activation energy Ea

log(k₂/k₁) = Ea/2.303R × (T₂ − T₁)/(T₁T₂) log(8.0/2.0) = Ea / (2.303 × 8.314) × 20 / (300 × 320) log 4 = Ea / 19.14 × 20 / 96000 0.6021 = Ea × 1.085 × 10⁻⁵ / 19.14 0.6021 = Ea × 5.668 × 10⁻⁷ Ea = 0.6021 / (5.668 × 10⁻⁷) Ea = 1.062 × 10⁶ J mol⁻¹ ... (recalculating carefully) Actually: 20/96000 = 2.083 × 10⁻⁴ 0.6021 = Ea × 2.083 × 10⁻⁴ / 19.14 0.6021 = Ea × 1.088 × 10⁻⁵ Ea = 0.6021 / (1.088 × 10⁻⁵) Ea = 55,340 J mol⁻¹ ≈ 55.3 kJ mol⁻¹

8. Important Board Questions with Model Answers

One-Mark Questions

Q1. Define order of a reaction. Order is the sum of the powers of concentration terms in the experimentally determined rate law equation.

Q2. Write the unit of rate constant for a first-order reaction. s⁻¹ (or min⁻¹). Unit of k for nth order = mol^(1−n) L^(n−1) s⁻¹.

Q3. State one example of a pseudo first-order reaction. Hydrolysis of methyl acetate (ester) in excess dilute HCl — water is in large excess so its concentration is effectively constant.

Q4. What happens to the rate constant when temperature increases? Rate constant k increases exponentially with temperature, as described by the Arrhenius equation k = Ae^(−Ea/RT).

Q5. How does a catalyst increase the rate of a reaction? A catalyst provides an alternate reaction pathway with lower activation energy, so more collisions have sufficient energy to react.


Short-Answer Questions (2–3 marks)

Q1. Distinguish order and molecularity.
Answer
FeatureOrderMolecularity
DefinitionSum of powers in experimental rate lawNumber of reacting species in elementary step
DeterminationExperimentallyTheoretically from mechanism
ValuesCan be 0, 1, 2, or fractionalOnly positive integers (1, 2, 3)
Change with conditionsCan changeCannot change
Q2. Derive the half-life of a first-order reaction.
Answer

For first order: k = (2.303/t) log([A]₀/[A]t)

At t = t½, [A]t = [A]₀/2:

k = (2.303/t½) log([A]₀/([A]₀/2)) k = (2.303/t½) log 2 k = (2.303/t½) × 0.3010 k = 0.693/t½ Therefore: t½ = 0.693/k

Key feature: t½ is independent of initial concentration for first-order reactions.

Q3. State the Arrhenius equation and explain what each term represents.
Answer

k = A·e^(−Ea/RT)

  • k = rate constant
  • A = frequency/pre-exponential factor (accounts for collision frequency and orientation)
  • Ea = activation energy (energy barrier in J mol⁻¹)
  • R = gas constant (8.314 J mol⁻¹ K⁻¹)
  • T = absolute temperature in Kelvin

The fraction of molecules with energy ≥ Ea is e^(−Ea/RT), which increases with T.


Long-Answer Questions (5 marks)

Q1. Derive the integrated rate equation for a first-order reaction. Sketch the graph and derive half-life.
Answer

For reaction A → products, rate = −d[A]/dt = k[A]

Separating variables: d[A]/[A] = −k dt

Integrating between limits [A]₀ at t=0 and [A]t at t:

∫([A]₀ to [A]t) d[A]/[A] = −k ∫(0 to t) dt ln[A]t − ln[A]₀ = −kt ln([A]t/[A]₀) = −kt [A]t = [A]₀ · e^(−kt)

In log form: log[A]t = log[A]₀ − kt/2.303

Graph: Plot log[A] vs t → straight line, slope = −k/2.303, y-intercept = log[A]₀

Half-life: When [A]t = [A]₀/2:

k·t½ = ln([A]₀/([A]₀/2)) = ln 2 = 0.693 t½ = 0.693/k
Q2. Explain how collision theory accounts for the effect of temperature and catalyst on reaction rate.
Answer

According to collision theory, reaction occurs only when molecules collide with (a) energy ≥ Ea and (b) correct orientation.

Effect of temperature: The Maxwell-Boltzmann distribution shows that at higher temperature, more molecules possess energy ≥ Ea. The fraction of effective molecules = e^(−Ea/RT). Since T appears in the exponent, even a small rise in T greatly increases this fraction, hence the dramatic increase in rate.

Effect of catalyst: A catalyst provides an alternate pathway with lower Ea'. The fraction of effective collisions on the new path = e^(−Ea'/RT) >> e^(−Ea/RT) since Ea' < Ea. Therefore more collisions are effective, and the rate increases. The activation energy itself, not the temperature, has changed.


Common Mistakes

  • Assuming order equals stoichiometric coefficient without experimental evidence
  • Thinking first-order half-life depends on initial concentration (it does not)
  • Mixing ln (natural log) and log (base 10) — the factor 2.303 converts between them
  • Stating that a catalyst increases equilibrium yield (it does not — only speeds up reaching equilibrium)
  • Using temperature in Celsius instead of Kelvin in the Arrhenius equation

Revision Checklist

  • [ ] Rate, rate law, order, and molecularity definitions
  • [ ] Determining order from experimental data tables
  • [ ] Zero-order and first-order integrated equations
  • [ ] First-order half-life derivation and numerical application
  • [ ] Pseudo first-order — definition and example
  • [ ] Arrhenius equation terms and logarithmic form
  • [ ] Finding Ea from two temperature rate constants
  • [ ] Collision theory — effective collision criteria
  • [ ] Effect of catalyst on activation energy and equilibrium

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Frequently Asked Questions

Rate of reaction is the speed at which reactants are converted to products at a given moment (changes with concentration). Rate constant k is a constant at fixed temperature — it does not depend on concentration but changes with temperature (Arrhenius equation).