A second-order reaction has a half-life of 20 minutes at an initial concentration of 0.1 M. What will be the half-life when the initial concentration is 0.05 M?
Correct: A
For a second-order reaction, t₁/₂ is inversely proportional to the initial concentration. Since [A]₀ is halved, t₁/₂ doubles. Original t₁/₂ = 1/(k·0.1) = 20 min. New t₁/₂ = 1/(k·0.05) = 40 min.