alleryStart learning →Which expression gives the half-life of a first order reaction?
At(1/2) = 1/(k[A]0)
Bt(1/2) = [A]0/(2k)
Ct(1/2) = 2k/[A]0
Dt(1/2) = 0.693/k
Answer & Solution
Correct answer: D. t(1/2) = 0.693/k
1. Start from ln([A]0/[A]) = kt and set [A] equal to half of [A]0.
2. The ratio inside the logarithm becomes 2, and ln 2 is 0.693.
3. So t(1/2) = 0.693/k for a first order reaction.
4. Note that [A]0 has cancelled out, which is why this half-life does not depend on the starting concentration.
5. The forms 1/(k[A]0) and [A]0/(2k) belong to second order and zero order respectively.
_Source: OpenStax Chemistry (CC BY 4.0), Ch 12 "Kinetics", section 12.4 Integrated Rate Laws_
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