Question

Prove that in case of first order reaction t₉₉.₉% = 10 t₁/₂

Hint:

A useful shortcut for first-order reactions: t₉₀% ≈ 3.33 t₁/₂, t₉₉% ≈ 2 × t₉₀%, and t₉₉.₉% ≈ 3 × t₉₀%. Since 10ⁿ = n, the time for "n-nines" completion increases linearly.

Answer 1

Concept: For a first-order reaction, the integrated rate equation relates the rate constant (k), time (t), initial concentration ([A]₀), and final concentration ([A]): k = (2.303)/(t) ([A]₀)/([A])
At half-life, the concentration becomes half of the initial value, i.e., [A] = ([A]₀)/(2). k = 2.303t₁/₂ ([A]₀)/([A]₀/2) = 2.303t₁/₂ 2 t₁/₂ = (2.303 × 0.3010)/(k) = (0.693)/(k) ·s (1)
For 99.9% completion, the amount of reactant reacted is 99.9% of [A]₀. Remaining concentration [A] = [A]₀ - 0.999[A]₀ = 0.001[A]₀ = 10⁻³[A]₀. k = 2.303t₉₉.₉% [A]₀10⁻³[A]₀ t₉₉.₉% = (2.303)/(k) 10³ = (2.303)/(k) × 3 t₉₉.₉% = (6.909)/(k) ·s (2)
Divide equation (2) by equation (1): t₉₉.₉%t₁/₂ = (6.909/k)/(0.693/k) ≈ 10 t₉₉.₉% = 10 t₁/₂ Hence proved.