Chapter 4 of Class 12 Chemistry — Chemical Kinetics — answers the question “How fast does a reaction go?” While thermodynamics tells us IF a reaction will happen, kinetics tells us HOW FAST. This chapter is crucial for 5-7 marks in Boards and is heavily tested in competitive exams. Focus on the integrated rate equations and Arrhenius equation numericals.
Key Concepts
Rate of a Chemical Reaction
Rate = −(1/a)(d[A]/dt) = −(1/b)(d[B]/dt) = (1/c)(d[C]/dt) = (1/d)(d[D]/dt)
Average rate: Δ[concentration] / Δt (over a time interval)
Instantaneous rate: d[concentration]/dt (at a specific moment — slope of tangent)
Factors Affecting Rate of Reaction
- Concentration: Higher concentration → more collisions → faster rate
- Temperature: 10°C rise ≈ doubles the rate (rough rule)
- Catalyst: Lowers activation energy → faster rate
- Nature of reactants: Ionic reactions are fast; covalent bond-breaking is slow
- Surface area: Larger surface area → faster rate (powdered vs lump)
Rate Law and Order of Reaction
Where k = rate constant, x = order w.r.t. A, y = order w.r.t. B
Overall order = x + y
Important: Order is determined experimentally, NOT from the balanced equation!
Molecularity IS determined from the elementary step.
Order vs Molecularity
| Property | Order | Molecularity |
|---|---|---|
| Determination | Experimental | Theoretical (from mechanism) |
| Value | Can be 0, fraction, integer | Always positive integer (1, 2, 3) |
| Applies to | Overall reaction | Elementary step only |
| Can be zero? | Yes | No |
Integrated Rate Equations
Zero Order Reaction
k = [A]₀ / t (when reaction is complete)
t₁/₂ = [A]₀ / 2k (half-life depends on initial concentration)
Unit of k: mol L⁻¹ s⁻¹
Graph: [A] vs t → straight line with slope = −k
First Order Reaction
Or: ln[A] = ln[A]₀ − kt
t₁/₂ = 0.693/k (half-life is INDEPENDENT of initial concentration!)
Unit of k: s⁻¹ (or min⁻¹, time⁻¹)
Graph: ln[A] vs t → straight line with slope = −k
Pseudo First Order Reaction
A reaction that is actually second order but behaves as first order because one reactant is in large excess.
- Hydrolysis of ester: CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH (water in excess)
- Inversion of sugar: C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O₆ + C₆H₁₂O₆ (water in excess)
Collision Theory
Where: Z_AB = collision frequency, Ea = activation energy, P = probability/steric factor
For a reaction to occur, collisions must be:
1. Energetically sufficient (energy ≥ Ea)
2. Properly oriented (correct geometry)
Arrhenius Equation
Taking log: log k = log A − Ea/(2.303RT)
For two temperatures:
log(k₂/k₁) = (Ea/2.303R) × [(T₂ − T₁)/(T₁T₂)]
Where: A = frequency factor (pre-exponential), Ea = activation energy (J/mol), R = 8.314 J/mol·K
Effect of Catalyst
- Catalyst provides an alternative pathway with lower activation energy
- Does NOT change ΔH or ΔG of the reaction
- Does NOT shift equilibrium — it speeds up both forward and reverse reactions equally
- It only helps reach equilibrium faster
Important Definitions
| Term | Definition |
|---|---|
| Rate of Reaction | Change in concentration of reactant or product per unit time |
| Rate Constant (k) | Proportionality constant in the rate law; rate when all concentrations are unity |
| Order of Reaction | Sum of powers of concentration terms in the rate law (experimental) |
| Molecularity | Number of molecules/atoms/ions taking part in an elementary step |
| Half-life (t₁/₂) | Time for concentration of reactant to fall to half its initial value |
| Activation Energy (Ea) | Minimum energy that colliding molecules must have for reaction to occur |
| Catalyst | Substance that increases rate by lowering Ea without being consumed |
Solved Examples — NCERT Based
Example 1: First Order Rate Constant
Q: A first order reaction has a rate constant of 1.15 × 10⁻³ s⁻¹. How long will 5 g of the reactant take to reduce to 3 g?
Solution:
k = (2.303/t) log([A]₀/[A])
1.15 × 10⁻³ = (2.303/t) log(5/3)
1.15 × 10⁻³ = (2.303/t) × 0.2219
t = (2.303 × 0.2219) / (1.15 × 10⁻³)
t = 0.5113 / 1.15 × 10⁻³ = 444.6 s ≈ 7.4 min
Example 2: Half-life of First Order Reaction
Q: The half-life of a first order reaction is 60 minutes. What percentage of the reactant will be left after 3 hours?
Solution:
3 hours = 180 minutes = 3 half-lives
After each half-life, amount halves:
After t₁/₂: 50% | After 2t₁/₂: 25% | After 3t₁/₂: 12.5%
Example 3: Arrhenius Equation — Finding Ea
Q: The rate constant of a reaction at 300 K is 1.6 × 10⁻² and at 310 K is 3.2 × 10⁻². Calculate the activation energy.
Solution:
log(k₂/k₁) = (Ea/2.303R) × [(T₂ − T₁)/(T₁T₂)]
log(3.2 × 10⁻²/1.6 × 10⁻²) = (Ea/(2.303 × 8.314)) × [(310 − 300)/(300 × 310)]
log(2) = (Ea/19.147) × (10/93000)
0.301 = Ea × 5.226 × 10⁻⁶ / 19.147
0.301 = Ea × 2.73 × 10⁻⁷
Ea = 0.301 / 2.73 × 10⁻⁷ = 1.103 × 10⁶ J/mol ≈ 55.3 kJ/mol
Example 4: Determining Order from Data
Q: For the reaction 2A + B → C, the following data is given:
| Exp | [A] | [B] | Rate |
|---|---|---|---|
| 1 | 0.1 | 0.1 | 2 × 10⁻³ |
| 2 | 0.2 | 0.1 | 4 × 10⁻³ |
| 3 | 0.1 | 0.2 | 2 × 10⁻³ |
Find the order and rate law.
Solution:
Comparing Exp 1 & 2: [A] doubles, [B] same → rate doubles → order w.r.t. A = 1
Comparing Exp 1 & 3: [B] doubles, [A] same → rate unchanged → order w.r.t. B = 0
Rate law: Rate = k[A]¹[B]⁰ = k[A] → Overall order = 1
Important Questions for Board Exams
1 Mark Questions
- Define the order of a reaction.
- What is the unit of rate constant for a first order reaction?
- Write the half-life expression for a zero order reaction.
- What is activation energy?
- Give an example of a pseudo first order reaction.
2 Mark Questions
- Distinguish between order and molecularity.
- A first order reaction has k = 2 × 10⁻³ s⁻¹. Find the half-life.
- What is the effect of a catalyst on activation energy? Draw energy profile diagram.
- Explain why the rate of reaction increases with temperature.
3 Mark Questions
- Derive the integrated rate equation for a first order reaction.
- For a first order reaction, show that t₁/₂ = 0.693/k.
- State Arrhenius equation. How can Ea be determined graphically?
- The rate constant doubles when temperature increases from 300 K to 310 K. Calculate Ea.
5 Mark Questions
- What is meant by order and molecularity? Derive the integrated rate equation for first order reaction. Also derive the expression for half-life.
- Explain collision theory of chemical kinetics. What are the conditions for effective collisions? Write the Arrhenius equation and explain the significance of each term.
Quick Revision Points
- Rate = change in concentration per unit time; always positive
- Order: experimental; Molecularity: theoretical (elementary step only)
- Zero order: [A] = [A]₀ − kt, t₁/₂ = [A]₀/2k
- First order: k = (2.303/t) log([A]₀/[A]), t₁/₂ = 0.693/k (independent of [A]₀)
- Pseudo first order: 2nd order reaction with one reactant in large excess
- Arrhenius: k = Ae^(−Ea/RT); plot log k vs 1/T → straight line
- For two temperatures: log(k₂/k₁) = (Ea/2.303R)[(T₂−T₁)/(T₁T₂)]
- Catalyst lowers Ea but doesn’t change ΔH or equilibrium
- 10°C rise ≈ doubles the rate (temperature coefficient ≈ 2)
- All radioactive decays follow first order kinetics
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