Chemical Kinetics
'Thermodynamics says what CAN happen. Kinetics says what DOES happen — and how fast.'
1. Chapter Overview
Chemical kinetics is the study of REACTION RATES and the FACTORS that influence them. Topics include: RATE OF A REACTION (average vs instantaneous rate), RATE LAW and RATE CONSTANT, ORDER and MOLECULARITY of reactions, INTEGRATED RATE EQUATIONS (zero-order, first-order, second-order reactions), HALF-LIFE of reactions, PSEUDO-FIRST-ORDER reactions, TEMPERATURE DEPENDENCE of reaction rates (Arrhenius equation, activation energy), and COLLISION THEORY.
2. Rate of a Reaction
- Average rate: −Δ[R]/Δt or Δ[P]/Δt (over a time interval).
- Instantaneous rate: −d[R]/dt or d[P]/dt (at a specific instant). 'The slope of the concentration vs time graph.'
- For the reaction 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.
3. Rate Law and Order
- Rate law: Rate = k[A]^m[B]^n. k = rate constant.
- Order of reaction: m + n (sum of exponents). Determined EXPERIMENTALLY.
- Molecularity: Number of molecules colliding in the RATE-DETERMINING step. Theoretical minimum based on reaction mechanism. ALWAYS a positive integer (1, 2, or very rarely 3).
Order vs Molecularity
| Aspect | Order | Molecularity |
|---|---|---|
| Definition | Sum of exponents in rate law | Number of molecules in rate-determining step |
| Determined by | EXPERIMENT | Reaction MECHANISM |
| Can be zero? | YES | NO (at least 1) |
| Can be fractional? | YES | NO (always integer) |
| Can be negative? | YES (for product inhibition) | NO |
4. Integrated Rate Equations
Zero-Order Reaction
- Rate = k. [A] = [A]₀ − kt. t_½ = [A]₀/(2k).
- 'The rate is CONSTANT — independent of concentration. Examples: Decomposition of NH₃ on Pt surface, some enzyme-catalysed reactions.'
First-Order Reaction
- Rate = k[A]. [A] = [A]₀ e^(−kt). ln[A] = ln[A]₀ − kt.
- k = (2.303/t) log([A]₀/[A]). t_½ = 0.693/k.
- 'Half-life is CONSTANT for first-order reactions — it does NOT depend on initial concentration.'
Second-Order Reaction
- Rate = k[A]². 1/[A] = 1/[A]₀ + kt. t_½ = 1/(k[A]₀).
- 'Half-life INVERSELY PROPORTIONAL to initial concentration.'
5. Pseudo-First-Order Reaction
- A reaction that is ACTUALLY second-order but behaves like first-order because one reactant is in LARGE EXCESS.
- Example: Hydrolysis of ethyl acetate: CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH. Water is in excess — its concentration remains essentially constant. Rate = k'[ester] where k' = k[H₂O].
Worked Example 1
Problem: A first-order reaction has rate constant 2.0×10⁻³ s⁻¹. How long will it take for 90% completion? Solution: t = (2.303/k) log([A]₀/[A]) = (2.303/2×10⁻³) log(100/10) = 1151.5 × log(10) = 1151.5 × 1 = 1151.5 s ≈ 19.2 min.
6. Temperature Dependence — Arrhenius Equation
- k = A e^(−Ea/RT) — A = frequency factor, Ea = activation energy.
- ln k = ln A − Ea/(RT).
- Two-temperature form: log(k₂/k₁) = (Ea/2.303R)(1/T₁ − 1/T₂).
Activation Energy (Ea)
- 'The MINIMUM energy that colliding molecules must have for the reaction to occur.'
- Low Ea → Fast reaction. High Ea → Slow reaction.
- Catalysts PROVIDE an alternative path with LOWER activation energy.
Worked Example 2
Problem: The rate constant doubles when temperature increases from 300 K to 310 K. Find Ea. Solution: log(2) = (Ea/2.303×8.314)(1/300 − 1/310). 0.3010 = (Ea/19.147)(0.003333 − 0.003226) = (Ea/19.147)(0.000107). Ea = 0.3010 × 19.147 / 0.000107 = 5.763 / 0.000107 ≈ 53,860 J/mol ≈ 53.9 kJ/mol.
7. Collision Theory
- Rate = Z × f × P — where Z = collision frequency, f = fraction with energy > Ea (e^(−Ea/RT)), P = steric (orientation) factor.
- 'Not EVERY collision leads to reaction. The molecules must have SUFFICIENT ENERGY and the CORRECT ORIENTATION.'
8. Comparison Table: Zero, First, and Second Order Reactions
| Feature | Zero Order | First Order | Second Order |
|---|---|---|---|
| Rate law | Rate = k | Rate = k[A] | Rate = k[A]² |
| Integrated eqn | [A]=[A]₀−kt | ln[A]=ln[A]₀−kt | 1/[A]=1/[A]₀+kt |
| Half-life | [A]₀/(2k) | 0.693/k | 1/(k[A]₀) |
| t_½ vs [A]₀ | t_½ ∝ [A]₀ | t_½ CONSTANT | t_½ ∝ 1/[A]₀ |
| Unit of k | mol L⁻¹ s⁻¹ | s⁻¹ | L mol⁻¹ s⁻¹ |
| Straight line plot | [A] vs t (gradient = −k) | log[A] vs t (gradient = −k/2.303) | 1/[A] vs t (gradient = k) |
9. Common Mistakes
- Order vs molecularity: Order is EXPERIMENTAL. Molecularity is THEORETICAL. They may NOT be equal.
- Units of k: k depends on ORDER. For nth order reaction: units = (mol/L)^(1−n) s⁻¹.
- Half-life of first-order reaction: t_½ = 0.693/k — does NOT depend on initial concentration.
- Arrhenius equation: log(k₂/k₁) = (Ea/2.303R)(1/T₁ − 1/T₂). T must be in KELVIN. Ea in J/mol.
10. CBSE Exam Focus
- Rate of reaction — average and instantaneous rate
- Rate law and order — determining order experimentally
- Integrated rate equations — zero, first order — numerical problems
- Half-life — first order (t_½ = 0.693/k)
- Arrhenius equation — activation energy, effect of temperature and catalyst
- Pseudo-first-order reactions
11. Self-Test
Q1: A first-order reaction has k = 1.0×10⁻³ s⁻¹. Find t_½ and the time for 75% completion. A1: t_½ = 0.693/10⁻³ = 693 s. For 75%, need 2 half-lives = 1386 s.
Q2: The half-life of a reaction is inversely proportional to initial concentration. What is the order? A2: Second order. (t_½ ∝ 1/[A]₀ for second order.)
Q3: A zero-order reaction has k = 0.04 mol L⁻¹ s⁻¹. Initial concentration = 1 M. Find t_½ and time for 90% completion. A3: t_½ = [A]₀/(2k) = 1/(0.08) = 12.5 s. For 90%: [A] = 0.1 M. t = ([A]₀−[A])/k = (1−0.1)/0.04 = 0.9/0.04 = 22.5 s.
Q4: At 300 K, k = 2.0×10⁻³ s⁻¹. At 400 K, k = 8.0×10⁻³ s⁻¹. Find Ea. A4: log(8/2) = (Ea/2.303×8.314)(1/300 − 1/400). log(4) = 0.6021. 0.6021 = (Ea/19.147)(0.003333−0.0025) = (Ea/19.147)(0.000833). Ea = 0.6021×19.147/0.000833 = 11.53/0.000833 ≈ 13,840 J/mol ≈ 13.8 kJ/mol.
Q5: For the reaction A + B → products, doubling [A] doubles the rate. Doubling [B] quadruples the rate. Write the rate law. A5: Rate = k[A][B]². (Order = 1+2 = 3. Third order overall.)
12. Conclusion
Chemical kinetics is about SPEED and MECHANISM:
- RATE: 'How fast does the reaction proceed? Measured by the change in concentration over time.'
- ORDER: 'What is the relationship between concentration and rate? Determined by EXPERIMENT.'
- HALF-LIFE: 'How long until half the reactant is consumed? For first-order reactions, it NEVER changes.'
- ARRHENIUS: 'Temperature speeds up reactions. Activation energy is the ENERGY BARRIER that reactions must overcome.'
'Kinetics tells us WHY some reactions that are thermodynamically favourable (diamonds → graphite) take MILLIONS of years — the activation energy is TOO HIGH.'
