Assam ASSEB HS Chemistry study materials
Assam ASSEB HS Chemistry study resources

Assam ASSEB HS Chemistry: Chemical Kinetics

Assam ASSEB HS Chemistry · Work through your exam, one topic at a time.

All guide sections

Chemical Kinetics

What this domain tests

Rate expressions, order versus molecularity, integrated rate laws, half-life, the Arrhenius equation, collision theory and catalysis.

Core rules

  • Rate = k[A]^x[B]^y; the order x + y comes from experiment, never directly from stoichiometric coefficients.
  • Units of k are mol^(1−n) L^(n−1) s⁻¹, so first order is s⁻¹ and second order is L mol⁻¹ s⁻¹.
  • Zero order: [A] = [A]₀ − kt, t½ = [A]₀/2k. First order: ln([A]₀/[A]) = kt, t½ = 0.693/k. Second order: 1/[A] − 1/[A]₀ = kt.
  • Arrhenius: k = A e^(−Ea/RT), so log(k₂/k₁) = (Ea/2.303R)(1/T₁ − 1/T₂).
  • Collision theory: a collision is effective only if E ≥ Ea and the orientation is correct; raising T sharply increases the fraction of molecules with E ≥ Ea.
  • A catalyst lowers Ea and increases both k_forward and k_reverse equally; it does not change ΔH or the equilibrium constant.

Worked example

A first-order reaction has t½ = 10 min. Then k = 0.693/10 = 0.0693 min⁻¹. For 90% decomposition, [A]/[A]₀ = 0.1, so t = (2.303/k) log(100/10) = (2.303/0.0693) × 1 = 33.2 min. The same k applies throughout; only the concentration ratio changes.

Traps that cost marks

Rate is positive for product formation and negative for reactant consumption; signs must match the species. Molecularity is a small whole number, but order may be zero, fractional or negative. For zero-order reactions t½ depends on initial concentration, unlike first order. A catalyst does not increase the yield at equilibrium, only the speed of reaching it.