Which plot would be linear for a second-order reaction of A → products with rate = k[A]^2?

• A. Plot [A] vs t
• B. Plot ln[A] vs t
• C. Plot 1/[A] vs t
• D. Plot log[A] vs t

Answer: (C)

For a second-order reaction in A with rate = k[A]^2, the way A changes over time is governed by -d[A]/dt = k[A]^2. Rearrange and integrate: d[A]/[A]^2 = -k dt. The integral gives -1/[A] = -kt + C, so 1/[A] = kt + C'. Using the initial concentration [A]0 at t = 0, you get 1/[A] = kt + 1/[A]0. This is a straight line when you plot 1/[A] versus time, with slope k and intercept 1/[A]0. The other plotted relationships aren’t linear for this second-order process: [A] versus t follows a hyperbolic form [A] = [A]0/(1 + k t [A]0); ln[A] versus t would be linear only for a first-order reaction; and plotting log[A] versus t does not produce a straight line for this rate law.