For a second-order reaction with rate = k[A]^2 and [A]0 = 0.100 M, k = 0.200 M^-1 s^-1, what is t1/2?

• A. 25 s
• B. 100 s
• C. 50 s
• D. 5 s

Answer: (C)

For a second-order reaction where rate = k[A]^2, the half-life depends on the initial concentration. Start from the rate equation d[A]/dt = -k[A]^2 and integrate: ∫d[A]/[A]^2 = -∫k dt gives 1/[A] = kt + 1/[A]0. At half-life, [A] = [A]0/2, so 2/[A]0 = kt + 1/[A]0, which simplifies to kt = 1/[A]0. Therefore t1/2 = 1/(k[A]0).

Plug in the values: k[A]0 = 0.200 M^-1 s^-1 × 0.100 M = 0.020 s^-1, so t1/2 = 1 / 0.020 s^-1 = 50 s.