Half-Life of a Second-Order Reaction
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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Grade 12Grade 12 Chemistry
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Half-Life of a Second-Order Reaction explained
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
- = Half-life (s)
- = Rate constant in L/(mol·s) (L/(mol·s))
- = Initial concentration (M)
Missing one of these? Work it out first, then come back
- Half-life — Half-Life Decay, Half-Life and Decay Constant
- Rate constant in L/(mol·s) — Zero-Order Integrated Rate Law, Second-Order Integrated Rate Law
- Initial concentration — Dilution Equation (C1V1 = C2V2), Zero-Order Integrated Rate Law