For a second-order reaction, which plot yields a straight line?

• A. A vs time
• B. ln[A] vs time
• C. A^2 vs time
• D. 1/[A] vs time

Answer: (D)

For a second-order reaction, the rate law is -d[A]/dt = k[A]^2. This leads to a simple integrated form: 1/[A] = kt + 1/[A]0. That means if you plot the reciprocal of the concentration against time, you get a straight line with slope k and intercept 1/[A]0. This is why the correct plot is 1/[A] versus time.

The other plots don’t give a straight line for a second-order process because they reflect different mathematical forms. A vs time is not linear because [A] decreases in a non-linear way under a second-order rate law. ln[A] vs time would be linear for a first-order process, not second-order. Plotting [A]^2 vs time also won’t yield a constant slope, since the integrated relationship isn’t linear in [A]^2.