What is the integrated rate law for a first-order reaction?

• A. [A] = [A]0 e^(kt)
• B. ln[A] = ln[A]0 − kt
• C. 1/[A] = 1/[A]0 + kt
• D. [A] = [A]0 − kt

Answer: (B)

A first-order reaction has a rate that depends on the concentration of A, described by -d[A]/dt = k[A]. Integrating this relationship gives ln[A] = ln[A]0 − kt, which is the integrated rate law. This form shows the natural log of the concentration decreasing linearly with time, with slope −k, and it can also be written as A = [A]0 e^(−kt).

This is the best answer because the logarithmic form directly reflects the exponential decay characteristic of a first-order process. The other forms correspond to different scenarios: [A] = [A]0 − kt would imply a constant rate (zero-order), [A] = [A]0 e^(kt) would imply growth rather than decay, and 1/[A] = 1/[A]0 + kt describes a second-order process where 1/[A] grows linearly with time.