Problem 4 - Entrance Test

In a first-order reaction, the time required for 99% completion is

A. 100 × t₅₀%B. 4.60 × t₅₀%C. 2 × t₅₀%D. 20 × t₅₀%

Correct: B

For first-order reactions, t = (1/k) ln([A₀]/[A]). t₉₉% = (1/k) ln(100), t₅₀% = (1/k) ln(2). Thus t₉₉% ≈ 4.60 × t₅₀%.