Which statement correctly pairs zero- and first-order half-life expressions?

Which statement correctly pairs zero- and first-order half-life expressions?

• A. Zero-order: t1/2 = 1/(k[A]); First-order: t1/2 = ln2/k
• B. Zero-order: t1/2 = [A]0/(2k); First-order: t1/2 = ln2/k
• C. Zero-order: t1/2 = 1/(k[A]); First-order: t1/2 = [A0]/k
• D. Zero-order: t1/2 = 1/(k[A]); First-order: t1/2 = ln2/k

Answer: (A)

Half-life tells us how long it takes for the concentration to drop by half, and the way it changes with time depends on the reaction order. For a zero-order process, the concentration falls linearly with time: [A] = [A]0 − kt. Halving the amount means [A] = [A]0/2, which gives t1/2 = [A]0/(2k). For a first-order process, the concentration decays exponentially: [A] = [A]0 e^(−kt). Setting [A] = [A]0/2 leads to e^(−kt1/2) = 1/2, so t1/2 = (ln 2)/k. These are the standard pairings. Forms like 1/(k[A]) or [A0]/k correspond to other kinetics (e.g., second-order) and do not describe the first-order or zero-order half-lives.

Half-life tells us how long it takes for the concentration to drop by half, and the way it changes with time depends on the reaction order.

For a zero-order process, the concentration falls linearly with time: [A] = [A]0 − kt. Halving the amount means [A] = [A]0/2, which gives t1/2 = [A]0/(2k).

For a first-order process, the concentration decays exponentially: [A] = [A]0 e^(−kt). Setting [A] = [A]0/2 leads to e^(−kt1/2) = 1/2, so t1/2 = (ln 2)/k.

These are the standard pairings. Forms like 1/(k[A]) or [A0]/k correspond to other kinetics (e.g., second-order) and do not describe the first-order or zero-order half-lives.