Which expression describes the first-order plot that yields a straight line?

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

Answer: (A)

For a first-order reaction, the rate law is -d[A]/dt = k[A]. Separating variables and integrating gives ln[A] = -kt + ln[A]0. This means plotting ln[A] versus time produces a straight line with slope -k and intercept ln[A]0. The other expressions correspond to different plot linearizations: [A] = [A]0 - kt would linearize a zero-order process, 1/[A] = kt + 1/[A]0 linearizes a second-order process, and [A] = [A]0 e^{-kt} is the exponential decay form that does not give a straight line on a plot of [A] versus time. The expression ln[A] = -kt + ln[A]0 is the one that yields a straight-line first-order plot.