Reaction Half-Life Calculator
Calculate the half-life of a zero-, first- or second-order chemical reaction from the rate constant k and, where needed, the initial concentration. Results are shown in seconds and in your chosen time unit.
Find the Half-Life of a Zero-, First- or Second-Order Reaction
Reaction Half-Life Formula
Where:
- \(t_{1/2}\) — half-life, the time needed for the reactant concentration to drop to half of its initial value
- \(k\) — rate constant; its units depend on the order (see below)
- \([A]_0\) — initial concentration of the reactant (mol/L)
Each expression comes from integrating the rate law of the corresponding order. Notice that only the first-order half-life is independent of the starting concentration — this is the behaviour familiar from radioactive decay.
Reaction Half-Life – Calculation Example
Example: a first-order decomposition has a rate constant of \(k = 0.05\ \text{min}^{-1}\) — the calculator's default values.
\[ t_{1/2} = \frac{\ln 2}{k} = \frac{0.6931}{0.05\ \text{min}^{-1}} = 13.86\ \text{min} \]
Result: the half-life is 13.86 minutes (about 831.8 seconds). Because the reaction is first order, no initial concentration is required — every half-life interval halves whatever concentration remains.
How to Use the Calculator
- Open Reaction order and choose Zero order, First order or Second order. For a first-order reaction the initial-concentration field is hidden automatically.
- Enter the Rate constant (k). Make sure its units match the order: mol·L⁻¹·time⁻¹ for zero order, time⁻¹ for first order, L·mol⁻¹·time⁻¹ for second order.
- Select the Time unit of k (per second, per minute, per hour or per day). The half-life is reported in the same unit.
- For zero- or second-order reactions, enter the Initial concentration (A₀) and pick its unit (mol/L, mmol/L or µmol/L). The value is converted to mol/L internally, so k must be referenced to mol/L (e.g. L·mol⁻¹·min⁻¹, not L·mmol⁻¹·min⁻¹).
- Read the results: the Half-Life in seconds, the Half-Life in Your Time Unit, and the Calculation Details showing the formula used and the substitution step by step.
What Is the Half-Life of a Reaction?
The half-life \(t_{1/2}\) is the time it takes for the concentration of a reactant to fall to one half of its value at the start of the interval. It is one of the most intuitive ways to describe how fast a reaction proceeds: a short half-life means the reactant disappears quickly, a long one means it lingers.
Unlike radioactive decay, where the half-life is fixed, the half-life of a chemical reaction depends on the order of the reaction — that is, on how the rate depends on concentration. For a first-order reaction the rate is proportional to \([A]\), and the half-life comes out constant: \(t_{1/2} = \ln 2 / k\). Successive half-lives always halve what is left, so 75% of the reactant is gone after two half-lives, 87.5% after three, and so on. This is exactly the pattern our Radioactive Decay Calculator uses.
Zero- and second-order reactions behave differently. In a zero-order reaction the rate is constant regardless of concentration, so \(t_{1/2} = [A]_0/2k\): the half-life grows as the reactant is used up, and the reaction finishes completely at \(t = [A]_0/k\), exactly twice the first half-life. In a second-order reaction \(t_{1/2} = 1/(k[A]_0)\), so each successive half-life doubles — doubling the initial concentration halves the half-life.
| Order | Half-life | Units of k | Dependence on \([A]_0\) |
|---|---|---|---|
| Zero | \(t_{1/2} = [A]_0/(2k)\) | mol·L⁻¹·time⁻¹ | Proportional |
| First | \(t_{1/2} = \ln 2/k\) | time⁻¹ | None |
| Second | \(t_{1/2} = 1/(k[A]_0)\) | L·mol⁻¹·time⁻¹ | Inversely proportional |
Measuring how the half-life changes with concentration is in fact a classic experimental method for determining the order of a reaction: if \(t_{1/2}\) is constant the reaction is first order; if it scales with \([A]_0\) it is zero order; if it is inversely proportional to \([A]_0\) it is second order.
Assumptions, Limitations and Practical Tips
- Single reactant, constant temperature. The formulas describe an isolated reactant whose integrated rate law holds over the whole concentration range. The rate constant \(k\) itself depends strongly on temperature, as described by the Arrhenius and Van 't Hoff equation; if the temperature changes, so does the half-life.
- Unit consistency matters. The concentration you enter is converted to mol/L, so \(k\) must carry its concentration part in mol·L⁻¹ (for example L·mol⁻¹·s⁻¹). A constant quoted per mmol or per µmol must be rescaled before use.
- Real reactions can change order. Many reactions follow simple kinetics only over a limited concentration range; near completion, side reactions, equilibria or depletion of a second reactant can alter the apparent order, so the calculated half-life is best treated as valid near the stated \([A]_0\).
- Pseudo-orders are common. If one reactant is in large excess, its concentration stays nearly constant and the reaction may behave as first order in the tracked reactant even though the true molecularity is higher. The first-order formula still applies with the pseudo-first-order rate constant.
- Successive half-lives. For zero order the reaction goes to completion at \(t = [A]_0/k\); for first order the reactant never fully disappears, and for second order each successive half-life doubles — keep this in mind when extrapolating far beyond the first half-life.
Frequently Asked Questions
How do I calculate the half-life of a reaction by hand?
First identify the reaction order, then apply the matching formula: for zero order use \(t_{1/2} = [A]_0/(2k)\), for first order \(t_{1/2} = \ln 2 / k \approx 0.693/k\), and for second order \(t_{1/2} = 1/(k[A]_0)\). Check that the units of \(k\) match the order, substitute your numbers, and the result comes out in the time unit used in \(k\). For example, a second-order reaction with \(k = 0.4\ \text{L·mol}^{-1}\text{s}^{-1}\) and \([A]_0 = 0.5\ \text{mol/L}\) has \(t_{1/2} = 1/(0.4 \times 0.5) = 5\) seconds.
Why doesn't a first-order half-life depend on the initial concentration?
Because the rate is directly proportional to the concentration that remains. When you integrate the first-order rate law, \([A]_0\) cancels out, leaving \(t_{1/2} = \ln 2 / k\). Whatever amount you start with, the same fraction (50%) is consumed in each interval. This is why radioactive decay and many first-order chemical decompositions have a characteristic, constant half-life — and why the calculator hides the concentration field for first-order reactions.
Can I use a rate constant given in different time or concentration units?
Yes, but convert it first. To change the time unit of \(k\), multiply or divide accordingly: \(k\) in s⁻¹ equals \(k\) in min⁻¹ divided by 60. For concentration units, the calculator converts your \([A]_0\) to mol/L, so a second-order \(k\) quoted in L·mmol⁻¹·min⁻¹ must be multiplied by 1000 to become L·mol⁻¹·min⁻¹ before entering it. Mixing units (e.g. a \(k\) per mmol with a concentration in mol/L) is the most common source of wrong half-lives.
How can I tell if my reaction is zero, first, or second order?
A classic method is to measure the half-life at several different starting concentrations. If \(t_{1/2}\) stays constant, the reaction is first order. If \(t_{1/2}\) doubles when \([A]_0\) doubles, it is zero order (proportional). If \(t_{1/2}\) halves when \([A]_0\) doubles, it is second order (inversely proportional). You can also plot \([A]\), \(\ln[A]\), or \(1/[A]\) against time and see which gives a straight line.
What happens after the first half-life — do all reactions keep halving at the same rate?
No, only first-order reactions do. For first order, each successive half-life is the same length: 50% remains after one, 25% after two, 12.5% after three. For second order, each successive half-life doubles, so the reaction slows down as it proceeds. For zero order, the rate is constant and the reactant is completely exhausted at \(t = [A]_0/k\) — exactly two first half-lives — so there are no further half-lives at all.
Why does my calculated half-life not match my experimental data?
Common causes include using \(k\) with mismatched units, applying the wrong order, or extrapolating beyond the valid range. Real reactions often change apparent order near completion, at low concentration, or when a second reactant is depleted. Temperature also matters — \(k\) follows the Arrhenius equation, so a 10 °C change can shift \(k\) by a factor of two or more. If a reactant is in large excess, you may be observing a pseudo-first-order constant, which works fine in the first-order formula but hides the true molecularity.
Is the half-life of a chemical reaction the same as radioactive half-life?
Only for first-order reactions. Radioactive decay is inherently first order, so its half-life is constant and independent of the amount present. Chemical reactions can be zero or second order too, where the half-life depends on the starting concentration. The mathematics of successive halving (75% gone after two half-lives, 87.5% after three) applies to any first-order process, chemical or nuclear, but not to zero- or second-order kinetics.