The formula
- First-order half-life
t½ = ln 2 / k ≈ 0.693 / k- Amount left after a whole number of half-lives
[A] = [A]₀ · (1/2)^(t / t½)- First-order integrated rate law
[A] = [A]₀ · e^(-kt), or ln[A] = ln[A]₀ - kt- Half-life for zero and second order, for contrast
t½ = [A]₀ / 2k (zero order); t½ = 1 / (k[A]₀) (second order)
What the symbols mean
| Symbol | Meaning | Unit |
|---|---|---|
t½ | Half-life, the time to consume half of what is present | s (any time unit, matched to k) |
k | Rate constant, or the decay constant for a radioactive sample | s⁻¹ for first order |
[A] | Concentration or amount left at time t | M (or any amount, since only the ratio matters) |
[A]₀ | Concentration or amount at the start | M |
t | Elapsed time | s |
ln 2 | Natural logarithm of 2, the constant linking half-life and rate constant | 0.693 (dimensionless) |
e | Base of the natural logarithm, used in the integrated form | 2.718 (dimensionless) |
When it applies
- The process is first order. Only then is the half-life a fixed number; a zero-order half-life shrinks as the sample is consumed and a second-order one grows.
- Keep k and t in matched time units. A rate constant per second used against a time in minutes puts the exponent out by a factor of 60.
- Radioactive decay is always first order, so the same t½ = 0.693 / k applies with k written as the decay constant.
- For a whole number of half-lives, halve repeatedly rather than reaching for the exponential; the two agree and the halving is harder to get wrong.
Worked example
Problem. A compound decomposes by first-order kinetics with k = 0.0125 s⁻¹. Find its half-life, and the time for the sample to fall to one eighth of its starting concentration.
- The reaction is first order, so use t½ = 0.693 / k = 0.693 / 0.0125 s⁻¹.
- t½ = 55.44 s, which is 55.4 s to three significant figures.
- One eighth is (1/2)³, so three half-lives have to pass: t = 3 × 55.44 = 166.3 s, which is 166 s.
- Check with the integrated form: [A]/[A]₀ = e^(-kt) = e^(-0.0125 × 166.3) = e^(-2.079) = 0.125, which is exactly one eighth, so the two routes agree.
Answer. A half-life of 55.4 s, and 166 s to fall to one eighth of the starting concentration.
Common mistakes
- Using t½ = 0.693 / k on a reaction that is not first order. Check the order before reaching for this equation.
- Thinking two half-lives leave nothing. Each half-life removes half of what is still there, so three leave one eighth rather than none.
- Dividing the starting amount by the number of half-lives instead of halving repeatedly.
- Mixing time units between k and t. A rate constant in s⁻¹ demands a time in seconds, and no other combination works.
Related formulas
- Rate law:
rate = k[A]^m[B]^n - Arrhenius equation:
k = A·e^(-Ea / RT)