First-Order Half-Life

For a first-order reaction, the time to consume half the reactant is constant: t1/2 = ln 2 / k ≈ 0.693/k.

t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{k}

t1/2 = ln2/k = 0.693/k

What it means

The half-life is the most intuitive kinetic number: how long until half is gone. For first-order reactions it is a pure constant — independent of how much you started with — because a fixed fraction decays per unit time. This makes it the fingerprint of first-order behavior: radioactive decay, many drug clearances, and first-order decompositions all share this property. After n half-lives, (1/2)^n of the material remains, so roughly 7 half-lives take you below 1%. When to use it: quoting reaction speed in plain language, estimating how long a first-order process takes to near completion, or converting between k and t1/2 in decay problems.

Variables

Symbol	Name	Unit	Meaning
$t1/2$	Half-life	s (or min, h — matches k's time unit)	Time for the reactant concentration to fall to half its current value.
$k$	Rate constant	time⁻¹	The first-order rate constant; larger k means a shorter half-life.

Assumptions

The reaction is strictly first order — this formula fails for zero- and second-order kinetics.
Temperature is constant, so k does not drift.
The reverse reaction is negligible.

Worked example

Problem

A first-order decomposition has k = 0.0231 min⁻¹. Find its half-life.

Given:

First-order reaction with k = 0.0231 min⁻¹

Write the first-order half-life: t1/2 = ln 2 / k = 0.693 / k.
Substitute: t1/2 = 0.6931 / 0.0231 min⁻¹.
Divide: 0.6931 / 0.0231 = 30.004 min, which rounds to 30.0 min.
Sanity check: after 30.0 min, [A] = [A]0 × e^(−kt) = [A]0 × e^(−0.693) = 0.500[A]0 — exactly half, as required.

t1/2 = 30.0 min

Every 30.0 minutes half of the remaining reactant disappears — after 60 min one-quarter remains, after 90 min one-eighth, and so on.