alleryStart learning →Which expression gives the half-life of a simple second order reaction?
At(1/2) = 1/(k[A]0)
Bt(1/2) = 0.693/k
Ct(1/2) = [A]0/(2k)
Dt(1/2) = k/[A]0
Answer & Solution
Correct answer: A. t(1/2) = 1/(k[A]0)
1. Start from 1/[A] minus 1/[A]0 equal to kt and set [A] equal to half of [A]0.
2. Then 2/[A]0 minus 1/[A]0 equals k times t(1/2), so 1/[A]0 equals k times t(1/2).
3. Rearranged, t(1/2) = 1/(k[A]0).
4. The starting concentration survives here, so this half-life does depend on how much reactant you began with.
5. The form 0.693/k is the first order result and cannot be used for a second order reaction.
_Source: OpenStax Chemistry (CC BY 4.0), Ch 12 "Kinetics", section 12.4 Integrated Rate Laws_
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