What is the integrated rate law for a second-order reaction with rate = k[A]^2?

• A. 1/[A] = 1/[A]0 + kt
• B. [A] = [A]0 e^{-kt}
• C. ln[A] = -kt + ln[A]0
• D. 1/[A] = [A]0^{-1} - kt

Answer: (D)

For a second-order reaction in A, where the rate is governed by -d[A]/dt = k[A]^2, the integrated form shows that the reciprocal of the concentration changes linearly with time.

Starting from d[A]/[A]^2 = -k dt and integrating gives -1/[A] = -kt + C. Using the initial condition at t = 0, where [A] = [A]0, we find C = -1/[A]0. Rearranging leads to 1/[A] = 1/[A]0 + kt. This means a plot of 1/[A] versus t is a straight line with slope k and intercept 1/[A]0.

The other forms correspond to different orders: exponential decay and the ln[A] form arise from first-order kinetics, and a negative kt term would indicate the wrong sign for this second-order case.