Which expression represents the integrated rate law for a first-order reaction?

• A. 1/[A]t = kt + 1/[A0]
• B. [A]t = -kt + [A0]
• C. ln[A]t = -kt + ln[A0]
• D. ln[A] = -kt + ln[A0]

Answer: (D)

For a first-order reaction, the rate law is -d[A]/dt = k[A]. Integrating this differential equation gives a logarithmic relation between concentration and time: ln([A]t) = -kt + ln([A]0). This means the natural log of the concentration at time t decreases linearly with time, with slope -k and an intercept equal to the natural log of the initial concentration.

The expression ln[A] = -kt + ln[A0] is the same relation, just written with left-hand side understood as the natural log of the concentration at time t and right-hand side using the initial concentration. It directly embodies the key idea: a straight-line dependence of ln([A]) on time, with a slope of -k and an intercept of ln([A]0). This form also makes it easy to see the alternative common expression [A]t = [A]0 e^{-kt}, since exponentiating both sides yields the same decay behavior.

The other forms either describe a different order (such as a linear dependence [A]t = [A]0 - kt, which would imply a constant rate that’s not correct for first-order) or correspond to different integrated forms (like 1/[A]t = kt + 1/[A]0 for second-order kinetics).