Which expression is an integrated rate law for a first-order reaction?

Which expression is an integrated rate law for a first-order reaction?

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

Answer: (C)

For a first-order reaction, the rate law is -d[A]/dt = k[A]. Separating variables gives d[A]/[A] = -k dt, and integrating from the initial time to time t leads to ln([A]t) - ln([A]0) = -kt. This simplifies to ln([A]t) = ln([A]0) - kt, which is the integrated rate law for a first-order process. The form ln([A]t) = -kt + ln([A]0) directly reflects how the natural log of the concentration decays linearly with time. The other forms correspond to different situations: [A]t = [A]0 - kt is for zero-order kinetics; 1/[A]t = kt + 1/[A]0 is for second-order kinetics; and rate = k[A] is the differential rate law, not the integrated one.

For a first-order reaction, the rate law is -d[A]/dt = k[A]. Separating variables gives d[A]/[A] = -k dt, and integrating from the initial time to time t leads to ln([A]t) - ln([A]0) = -kt. This simplifies to ln([A]t) = ln([A]0) - kt, which is the integrated rate law for a first-order process. The form ln([A]t) = -kt + ln([A]0) directly reflects how the natural log of the concentration decays linearly with time.

The other forms correspond to different situations: [A]t = [A]0 - kt is for zero-order kinetics; 1/[A]t = kt + 1/[A]0 is for second-order kinetics; and rate = k[A] is the differential rate law, not the integrated one.