About the Integrated Rate Law Calculator
This integrated rate law calculator solves the three standard kinetics equations that link a reactant’s concentration to time. Pick the reaction order (zero, first or second), then choose what you want: the concentration remaining after a given time, the time needed to fall to a target concentration, or the rate constant k from two measured concentrations. Every answer comes with the half-life and the percentage of reactant left.
It is built for general chemistry and physical chemistry students working kinetics problems, and for anyone in the lab estimating how long a decomposition, a drug-degradation step or a first-order decay will take. The chart draws the full concentration–time curve so you can see the difference between the straight line of a zero-order reaction, the exponential decay of a first-order reaction and the long tail of a second-order one.
The equations assume a single reactant (or pseudo-order conditions where other reactants are in large excess), a constant temperature and a rate constant that does not change during the run. Use any time unit you like — seconds, minutes, hours or days — as long as k is expressed per that same unit.
How to use the integrated rate law calculator
- 1Choose the reaction order — it is usually given or found from a rate-law experiment.
- 2Choose whether you want the concentration, the time or the rate constant.
- 3Pick the time unit that matches your rate constant.
- 4Enter [A]₀ and the two remaining known values.
- 5Read the answer, half-life and percentage left, and check the curve on the chart.
Formula and method
Each integrated rate law comes from integrating the differential rate law −d[A]/dt = k[A]ⁿ for n = 0, 1 or 2. A zero-order reaction loses concentration at a constant rate, so [A] falls in a straight line. A first-order reaction decays exponentially, [A] = [A]₀e^(−kt). A second-order reaction gives a straight line when 1/[A] is plotted against time.
Rearranging the same equations gives the time to reach a target concentration or the rate constant from two measurements. Half-lives follow by setting [A] = [A]₀/2: t½ = [A]₀/(2k) for zero order, ln 2/k for first order and 1/(k[A]₀) for second order. Only the first-order half-life is independent of the starting concentration.
- [A]₀
- Initial concentration (M)
- [A]
- Concentration at time t (M)
- k
- Rate constant (units depend on the order)
- t
- Elapsed time
- t½
- Half-life
Worked examples
First-order concentration after 60 s
With k = 0.025 s⁻¹, [A] = 0.50 × e^(−0.025 × 60) = 0.50 × e^(−1.5) ≈ 0.1116 M. The half-life is ln 2 ÷ 0.025 ≈ 27.7 s, so after 60 s (about 2.2 half-lives) roughly 22.3% remains.
Zero-order time to fall from 1.00 M to 0.25 M
A zero-order reaction loses a fixed 0.015 M every second, so dropping 0.75 M takes 0.75 ÷ 0.015 = 50 s. The first half-life is 1.00 ÷ (2 × 0.015) ≈ 33.3 s.
Second-order rate constant from two concentrations
For a second-order reaction, k = (1/[A] − 1/[A]₀) ÷ t = (20 − 5) ÷ 300 = 0.050 M⁻¹·s⁻¹. The half-life from 0.20 M is 1 ÷ (0.050 × 0.20) = 100 s.
Frequently asked questions
How do I know which integrated rate law to use?+
Plot your data three ways: [A] vs t, ln[A] vs t and 1/[A] vs t. Whichever plot is a straight line tells you the order — zero, first or second respectively — and its slope gives k (with sign −k, −k or +k).
Why is the first-order half-life constant?+
Because t½ = ln 2 ÷ k contains no concentration term. A first-order reaction always takes the same time to lose half of what is left, which is why radioactive decay is described by a fixed half-life.
What are the units of the rate constant?+
They depend on the order so that the rate comes out in M per unit time: M·s⁻¹ for zero order, s⁻¹ for first order and M⁻¹·s⁻¹ for second order. The calculator shows the units for your chosen order and time unit.
What happens when a zero-order reaction runs out?+
The equation [A] = [A]₀ − kt would go negative, which is impossible. Real zero-order reactions stop at zero (or switch to a different order as the catalyst surface is no longer saturated), so the calculator clamps [A] at 0.
Can I use this for pseudo-first-order reactions?+
Yes. If a second reactant is in large excess its concentration barely changes, and the reaction behaves as first order with an observed constant k′ = k[B]. Enter k′ as the first-order rate constant.