In a second-order reaction with rate law rate = k[A]^2, the half-life is t1/2 = ?

• A. 1/(k[A]0)
• B. k[A]0/2
• C. [A]0/(2k)
• D. 2/[k[A]0]

Answer: (A)

For a second-order reaction where rate = k[A]^2, the rate equation is d[A]/dt = -k[A]^2. Integrating gives 1/[A] = kt + 1/[A]0. To find the half-life, set [A] = [A]0/2 and substitute:

1/([A]0/2) = kt1/2 + 1/[A]0

2/[A]0 = kt1/2 + 1/[A]0

kt1/2 = 1/[A]0

t1/2 = 1/(k[A]0)

So the half-life is 1/(k[A]0). This shows why the half-life for this second-order process depends on the initial concentration, unlike first-order reactions where t1/2 is constant.