Hong Life CalculatorsUnderstanding the reaction kinetics and half-life calculator — concept and principle
Chemical kinetics tells you how the concentration of a reactant falls over time. This calculator uses the integrated rate laws for zero-, first- and second-order reactions to give the concentration left after a chosen time and the half-life, the time needed for the concentration to drop to half. It is used for homework problems, drug elimination estimates, shelf-life questions and lab data analysis.
How it is calculated
- Zero order: [A] = [A]₀ − kt (never below 0); t½ = [A]₀ / (2k)
- First order: [A] = [A]₀ × e^(−kt); t½ = ln 2 / k
- Second order: 1/[A] = 1/[A]₀ + kt; t½ = 1 / (k[A]₀)
The units of k depend on the order: concentration/time for zero order, 1/time for first order, 1/(concentration × time) for second order. Use the same time unit for k and t, and the same concentration unit for k and [A]₀; the half-life comes out in that time unit.
The principle
The rate law says how the rate depends on concentration: rate = k[A]ⁿ. For n = 0 the rate is constant, so the concentration falls in a straight line. For n = 1 the rate is proportional to what is left, so the same fraction disappears in every equal time interval; that produces exponential decay and a constant half-life. For n = 2 the rate falls with the square of the concentration, so the reaction slows down sharply as the reactant is used up, and each successive half-life is twice as long as the one before. The integrated formulas come from solving d[A]/dt = −k[A]ⁿ.
Worked example
- Inputs: order = first, k = 0.05 per time unit, [A]₀ = 1, t = 10.
- kt = 0.05 × 10 = 0.5.
- [A] = 1 × e^(−0.5) = 0.6065, so about 60.7% remains.
- t½ = ln 2 ÷ 0.05 = 0.693147 ÷ 0.05 = 13.86 time units.
| Initial concentration [A]₀ | [A] at t |
|---|---|
| 0.25 | 0.1516326649 |
| 0.5 | 0.3032653299 |
| 0.75 | 0.4548979948 |
| 1 | 0.6065306597 |
| 1.25 | 0.7581633246 |
| 1.5 | 0.9097959896 |
| 2 | 1.213061319 |
How to read the result
For a first-order reaction the half-life does not depend on the starting amount: after 1, 2 and 3 half-lives 50%, 25% and 12.5% remain. For zero order the half-life grows with [A]₀; for second order it shrinks as [A]₀ grows. To find the order from data, check which plot is a straight line: [A] versus t (zero), ln[A] versus t (first), or 1/[A] versus t (second).
Common mistakes
- Mixing time units, such as k in per minute with t in seconds.
- Using the first-order half-life formula for a second-order reaction.
- Assuming the reaction order equals the coefficient in the balanced equation; order is found by experiment.
Frequently asked questions
What is a pseudo-first-order reaction?
When one reactant is in large excess, its concentration barely changes, so the rate depends on the other reactant alone and behaves as first order.
How many half-lives until a reaction is "complete"?
After about 7 half-lives less than 1% remains (0.5⁷ ≈ 0.8%) for a first-order process.
Why does a zero-order concentration stop at zero?
The straight-line law would turn negative after [A]₀/k, which is impossible, so the result is capped at 0.
Sources: P. Atkins and J. de Paula, Physical Chemistry, Oxford University Press (chemical kinetics) · IUPAC Compendium of Chemical Terminology (Gold Book), "order of reaction" and "half-life"