Rate Constant Calculator for A-Level Chemistry

Last updated: June 2026

Use this rate constant and half-life calculator to solve rate-law questions and zero-, first- or second-order half-life problems. The two ideas are kept in separate calculator cards so that the required inputs and units remain clear.

Key takeaway: The rate constant k links reaction rate to concentration. The half-life equation depends on reaction order: zero- and second-order half-lives require the initial concentration, while a first-order half-life does not.
How to use these calculators
  1. Use the Rate Constant Calculator when the rate equation, reaction rate and reactant concentrations are known.
  2. Select the experimentally determined order for each reactant. A zero-order concentration is automatically ignored.
  3. Use the Half-Life Calculator to choose zero-, first- or second-order kinetics and calculate either t1/2 or k.
  4. Enter the initial concentration when zero- or second-order kinetics is selected.
  5. Select a preset to fill an example, then press Calculate to see the equation, substitution and units.

What is the rate constant?

The rate constant, k, is the proportionality constant in a rate equation. At a fixed temperature, it connects the measured reaction rate to the concentrations that appear in the experimentally determined rate law.

rate = k[A]m[B]n

The powers m and n are reaction orders found from experimental data. They are not normally copied from the balanced chemical equation.

How to calculate k from a rate equation

Rearranging the rate equation gives:

k = rate / ([A]m[B]n)

A zero-order reactant contributes a factor of 1 because any non-zero concentration raised to the power zero equals 1. Its concentration is therefore not needed in the calculation.

How the units of k depend on overall order

The units of k must combine with the concentration terms to reproduce the units of reaction rate. If rate is measured in mol dm−3 s−1, then:

Overall order 0: mol dm−3 s−1
Overall order 1: s−1
Overall order 2: dm3 mol−1 s−1
Overall order 3: dm6 mol−2 s−1
Overall order 4: dm9 mol−3 s−1

What reaction order means

A zero-order reactant does not affect the rate when its concentration changes. A first-order reactant produces a directly proportional change, while a second-order reactant has a squared effect.

When several reactants appear in the rate law, the overall order is the sum of the individual orders.

Half-life equations for different reaction orders

Half-life is the time required for the concentration of a reactant to fall to half its initial value. The equation used depends on the order of a single-reactant rate law, rate = k[A]n.

Zero-order half-life

t1/2 = [A]0 / (2k)

A zero-order half-life is directly proportional to the initial concentration. Doubling [A]0 doubles the half-life when k is unchanged.

First-order half-life

t1/2 = 0.693 / k

A first-order half-life is independent of the initial concentration. Every successive halving therefore takes the same amount of time.

Second-order half-life

t1/2 = 1 / (k[A]0)

A second-order half-life is inversely proportional to the initial concentration. A larger [A]0 gives a shorter half-life when k is unchanged.

Keep the time unit consistent. For example, if k is expressed using seconds, the calculated half-life is in seconds. The concentration used by this calculator is mol dm−3.

Worked examples

Finding k for a zero-order rate law

If rate = k and the measured rate is 2.50 × 10−4 mol dm−3 s−1, then:

k = 2.50 × 10−4 mol dm−3 s−1

Finding k for a first-order rate law

When rate = k[A], [A] = 0.200 mol dm−3 and rate = 4.00 × 10−3 mol dm−3 s−1:

k = (4.00 × 10−3) / 0.200 = 2.00 × 10−2 s−1

Finding k for an overall second-order rate law

When rate = k[A][B], [A] = 0.100 mol dm−3, [B] = 0.200 mol dm−3 and rate = 6.00 × 10−4 mol dm−3 s−1:

k = (6.00 × 10−4) / (0.100 × 0.200) = 3.00 × 10−2 dm3 mol−1 s−1

Zero-order half-life

If [A]0 = 0.100 mol dm−3 and k = 5.00 × 10−3 mol dm−3 s−1:

t1/2 = 0.100 / (2 × 5.00 × 10−3) = 10.0 s

First-order half-life

t1/2 = 0.693 / 0.0500 = 13.9 s

Second-order half-life

If [A]0 = 0.200 mol dm−3 and k = 0.250 dm3 mol−1 s−1:

t1/2 = 1 / (0.250 × 0.200) = 20.0 s

Finding a first-order k from half-life

k = 0.693 / 120 = 5.78 × 10−3 s−1

Common exam traps

Do not obtain reaction orders from the balanced equation. Orders are normally determined experimentally.
Do not automatically give k the unit s−1. That unit applies only to an overall first-order reaction when time is measured in seconds.
Do not use t1/2 = 0.693 / k for every reaction. Zero- and second-order reactions use different equations and require [A]0.

Frequently asked questions

Can k be found from the balanced equation alone?

No. You need an experimentally determined rate equation together with rate and concentration data.

Why does k change with temperature?

Increasing temperature raises the fraction of particles able to overcome the activation energy, so the rate constant normally increases.

Does a catalyst change k?

Yes. A catalyst provides an alternative pathway with lower activation energy, increasing the rate constant at the same temperature.

Why is first-order half-life constant?

The first-order half-life equation contains only k, not [A]0. Each successive halving therefore takes the same time at a fixed temperature.

Why is initial concentration required for zero- and second-order half-life?

Their integrated rate equations contain [A]0. Changing the starting concentration therefore changes the time needed for the concentration to fall by half.

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