Q1(g). Half-life of a second-order reaction

Answer in brief

t½ = 1/k[A]₀: it depends on the initial concentration, inversely.

Key points

• Rate law for A → products: −d[A]/dt = k[A]².
• Separating and integrating: 1/[A] − 1/[A]₀ = kt.
• At t = t½, [A] = [A]₀/2: 2/[A]₀ − 1/[A]₀ = kt½ ⇒ t½ = 1/k[A]₀.
• So t½ is inversely proportional to [A]₀: with k = 0.50 L mol⁻¹ s⁻¹, 0.10 M gives 20 s and 0.20 M gives 10 s; successive half-lives double.
• Contrast: first order t½ = 0.693/k (independent); zero order [A]₀/2k; in general t½ ∝ [A]₀¹⁻ⁿ.