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Half-Life of a Second-Order Reaction

t1/2=1k [A]0t_{1/2} = \frac{1}{k\,[\mathrm{A}]_0}
L/(mol·s)
sig figs

Worked example: k = 0.200 L/(mol·s), 0.500 M → t_half = 10.0 s. Press Try an example to run it live, then adjust anything.

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Half-Life Decay
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See also

Second-Order Integrated Rate Law: Both use Rate constant in L/ and initial concentration.Half-Life and Decay Constant: Both use half life.Half-Life Decay: Both use half life.
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Half-Life of a Second-Order Reaction explained

[A]0t1/2k

Set [A] = [A]₀/2 in the second-order integrated law and the reciprocals collapse to t½ = 1/(k[A]₀). The consequence is counterintuitive if you are used to radioactive decay: the half-life is not a constant of the reaction but depends on where you start, and it doubles every time you halve the concentration. With k = 0.200 L/(mol·s) and [A]₀ = 0.500 M the first half-life is 1/(0.200 × 0.500) = 10.0 s; the next half, from 0.250 M down to 0.125 M, takes 20.0 s, then 40.0 s, and so on.

That lengthening tail is the signature of second-order kinetics and a genuinely useful diagnostic at the bench: measure successive half-lives and if they keep doubling, the reaction is second order; if they stay constant, first order; if they keep halving, zero order. It also explains why the last traces of a dimerising impurity are so stubborn to remove: the reaction that cleans it up slows down quadratically as the impurity thins out.

Half-Life of a Second-Order Reaction formula

t1/2=1k [A]0t_{1/2} = \frac{1}{k\,[\mathrm{A}]_0}
Where
  • t1/2t_{1/2}= Half-life (s)
  • kk= Rate constant in L/(mol·s) (L/(mol·s))
  • [A]0[\mathrm{A}]_0= Initial concentration (M)
In this toolkit

The integrated rate laws

5 formulasZero-, first- and second-order concentration-versus-time laws and their half-lives — and how to read the order off a straight line.

Revised 31 October 2026
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formula.expert. (2026, October 31). Half-Life of a Second-Order Reaction Formula & Calculator. https://formula.expert/solvers/second-order-half-life

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