Q1(h). Half-life of a first order reaction

Answer in brief

t₁/₂ = ln 2/k = 0.693/k: [A]₀ cancels, so the half-life does not depend on the initial concentration.

Key points

• Rate law: −d[A]/dt = k[A]; separate and integrate from [A]₀ at t = 0.
• ln([A]₀/[A]) = kt, i.e. [A] = [A]₀e⁻ᵏᵗ.
• At t = t₁/₂, [A] = [A]₀/2: ln 2 = kt₁/₂ ⇒ t₁/₂ = 0.693/k; [A]₀ cancels.
• Every half-life is equal; for order n, t₁/₂ ∝ 1/[A]₀ⁿ⁻¹: only n = 1 gives a zero power.
• Examples: radioactive decay (¹⁴C, t₁/₂ = 5730 y), decomposition of N₂O₅, catalysed decomposition of H₂O₂.