What is the half-life expression for a first-order reaction?

• A. t1/2 = 1/(k[A])
• B. t1/2 = [A]0/(2k)
• C. t1/2 = k/[A0]
• D. t1/2 = ln2 / k

Answer: (D)

For a first-order reaction, the rate law is proportional to the concentration, so -d[A]/dt = k[A]. Integrating gives [A] = [A]0 e^{-kt}. The half-life is the time it takes for the concentration to drop to half of its initial value: [A] = [A]0/2. Substituting and solving 1/2 = e^{-kt1/2} yields t1/2 = ln(2)/k, which is often written as t1/2 = ln2 / k. This shows the half-life depends only on the rate constant k, not on the starting concentration. For reference, zero-order half-life scales with [A]0, while second-order half-life scales with 1/[A]0, so the ln2/k form uniquely describes the first-order case.