For a second-order reaction, which equation represents the straight-line form on a plot of 1/[A] versus time?

• A. [A] = [A]0 - kt
• B. ln[A] = -kt + ln[A]0
• C. [A] = [A]0 e^{-kt}
• D. 1/[A] = kt + 1/[A0]

Answer: (D)

For a second-order reaction where the rate depends on the square of the concentration, -d[A]/dt = k[A]^2. Rearranging and integrating gives 1/[A] = kt + 1/[A]0. This means when you plot 1/[A] against time, you get a straight line with slope k and y-intercept 1/[A]0. That’s why the linear form for this plot is 1/[A] = kt + 1/[A]0. The other common relationships correspond to different plots (first-order or zero-order) and would not produce a straight line for 1/[A] versus time.