Last updated: 2026-06-16

A second-order reaction has a rate proportional to the square of one reactant's concentration, or to the product of two concentrations. The single-reactant case, rate $= k[A]^2$, is the one handled here. This calculator applies the second-order integrated rate law to find the concentration left after a given time and reports the initial half-life. Second-order kinetics are typical of bimolecular steps, where two molecules must collide for a reaction to occur.

The integrated rate law

Integrating $-\dfrac{d[A]}{dt} = k[A]^2$ gives a reciprocal straight-line form:

$\frac{1}{[A]} = \frac{1}{[A]_0} + kt \qquad\Longrightarrow\qquad [A] = \frac{1}{\,1/[A]_0 + kt\,}$

A plot of $[A]^{-1}$ against time is linear with slope $k$, which distinguishes second-order behaviour from the $\ln[A]$ line of a first-order reaction. The rate constant carries units of $\mathrm{M^{-1}s^{-1}}$.

A lengthening half-life

Setting $[A] = [A]_0/2$ gives a half-life that depends on the starting concentration:

$t_{1/2} = \frac{1}{k[A]_0}$

As the reactant is consumed, the concentration entering each new interval is smaller, so every successive half-life is twice as long as the one before — the opposite of the constant first-order half-life.

Worked example

With $[A]_0 = 1.0\ \mathrm{M}$ and $k = 0.05\ \mathrm{M^{-1}s^{-1}}$, after 60 seconds:

The initial half-life is $t_{1/2} = 1/(0.05 \times 1.0) = 20\ \mathrm{s}$.

Why the reciprocal appears

Because the rate falls with the square of concentration, it drops faster than first order at first but then lingers at low concentrations, since the few remaining molecules rarely meet. Integrating that $[A]^2$ dependence produces the reciprocal relationship rather than a logarithm. Compare the First-Order Integrated Rate Law Calculator and the Zero-Order Integrated Rate Law Calculator to see how the order reshapes the decay, and use the Average Reaction Rate Calculator to read an overall rate from concentration data.