For a second-order rate law, which plot is a straight line?

For a second-order rate law, which plot is a straight line?

• A. Concentration versus time
• B. Ln concentration versus time
• C. 1 over concentration versus time
• D. Concentration squared versus time

Answer: (C)

For a second-order reaction, the relationship between concentration and time becomes linear when you plot the reciprocal of the concentration against time. Starting from rate = k[A]^2, you can separate variables and integrate: -d[A]/dt = k[A]^2 leads to ∫d[A]/[A]^2 = -∫k dt, which gives 1/[A] = kt + 1/[A]0. This means the plot of 1/[A] versus time is a straight line with slope k and intercept 1/[A]0. The other plots don’t produce a straight line for a second-order process: [A] versus time is curved, and ln[A] versus time would only be linear for a first-order reaction. Therefore, the straight-line plot for a second-order rate law is 1/[A] versus time.

For a second-order reaction, the relationship between concentration and time becomes linear when you plot the reciprocal of the concentration against time. Starting from rate = k[A]^2, you can separate variables and integrate: -d[A]/dt = k[A]^2 leads to ∫d[A]/[A]^2 = -∫k dt, which gives 1/[A] = kt + 1/[A]0. This means the plot of 1/[A] versus time is a straight line with slope k and intercept 1/[A]0. The other plots don’t produce a straight line for a second-order process: [A] versus time is curved, and ln[A] versus time would only be linear for a first-order reaction. Therefore, the straight-line plot for a second-order rate law is 1/[A] versus time.