Which plot correctly linearizes a second-order rate law data?

Which plot correctly linearizes a second-order rate law data?

• A. [A] versus t
• B. ln[A] versus t
• C. 1/[A] versus t
• D. A^2 versus t

Answer: (C)

For a reaction that is second order in A, the rate law is rate = k[A]^2. The integrated form of this differential equation is 1/[A] = kt + 1/[A]0, meaning that plotting 1/[A] against time gives a straight line with slope k and intercept 1/[A]0. This makes 1/[A] versus time the correct linearization, because the math of a second-order decay converts the inverse concentration into a linear function of time. Plotting [A] versus time isn’t linear for a second-order process, since [A] follows [A]0/[1 + k[A]0 t], which curves downward. Plotting ln[A] versus time would be linear only for a first-order process. Plotting A^2 versus time isn’t a standard linear form for this rate law, so it wouldn’t yield a straight line.

For a reaction that is second order in A, the rate law is rate = k[A]^2. The integrated form of this differential equation is 1/[A] = kt + 1/[A]0, meaning that plotting 1/[A] against time gives a straight line with slope k and intercept 1/[A]0. This makes 1/[A] versus time the correct linearization, because the math of a second-order decay converts the inverse concentration into a linear function of time.

Plotting [A] versus time isn’t linear for a second-order process, since [A] follows [A]0/[1 + k[A]0 t], which curves downward. Plotting ln[A] versus time would be linear only for a first-order process. Plotting A^2 versus time isn’t a standard linear form for this rate law, so it wouldn’t yield a straight line.