First-Order Half-Life Calculator
For a first-order chemical reaction or radioactive decay, the half-life is the time taken for the concentration or amount to decrease to half its current value. The half-life is constant and is related to the rate constant k by t(1/2) = ln(2)/k = 0.6931/k. This calculator works in both directions: enter k to find t(1/2), or enter t(1/2) to find k. It also calculates the remaining fraction of reactant after a given number of half-lives.
First-order half-life formula
t(1/2) = ln(2) / k = 0.6931 / k k = ln(2) / t(1/2) [A](t) = [A]0 * exp(-k*t) = [A]0 * (1/2)^(t / t(1/2)) remaining % = (1/2)^n * 100
Where n is the number of half-lives elapsed. The half-life is independent of initial concentration for first-order reactions.
Applications of first-order kinetics
- Radioactive decay: carbon-14 has a half-life of 5,730 years; used in radiocarbon dating.
- Drug pharmacokinetics: most drugs are eliminated from the body by first-order processes.
- Thermal decomposition: many unimolecular decomposition reactions are first-order.
- After 10 half-lives, less than 0.1% of the original substance remains.
- Time to reach equilibrium is approximately 4 to 5 half-lives for most practical purposes.
Frequently asked questions
What is half-life in a first-order reaction?
The half-life (t(1/2)) is the time required for the concentration of a reactant to decrease to half its initial value. For a first-order reaction, the half-life is constant and independent of the initial concentration: t(1/2) = ln(2)/k = 0.6931/k.
Why is the half-life constant for first-order reactions?
For first-order reactions, the integrated rate law gives [A] = [A]0 * exp(-kt). Setting [A] = [A]0/2 and solving for t gives t(1/2) = ln(2)/k, which contains no [A]0 term. The half-life is therefore independent of initial concentration.
Does this apply to radioactive decay?
Yes. Radioactive decay is always first-order. The decay constant lambda plays the same role as k. The activity halves every t(1/2) = 0.6931/lambda, regardless of how much material is present.
How many half-lives to reduce to 1% of original?
Each half-life reduces the amount by half. After n half-lives, the remaining fraction is (1/2)^n. To reach 1% (0.01), solve (1/2)^n = 0.01, giving n = log(0.01)/log(0.5) = approximately 6.64 half-lives.
What is the relationship between k and t(1/2)?
k = ln(2) / t(1/2) = 0.6931 / t(1/2). A longer half-life means a smaller rate constant (slower reaction). A shorter half-life means a larger rate constant (faster reaction).
Official sources
- IUPAC Gold Book: Half-life definition.
- NIST Chemical Kinetics Database: NIST kinetics data.
Reviewed by the CalculatorHub team, edited by James Graham, 15 June 2026. See our methodology.