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Question
Derive an expression for the relation between half-life and rate constant for first-order reaction.
Solution
The integrated rate law for the first-order reaction is
k = `2.303/t * log_10 [A]_0/[A]_t`
Where [A]0 is the initial concentration of reactant at t = 0. It falls to [A]t at time t after the start of the reaction.
The time required for [A]0 to become `[A]_0/2` is denoted as t1/2 or [A]t = `[A]_0/2` at t = t1/2
Putting this condition in the integrated rate law we write
k = `2.303/"t"_(1//2) log_10 [A]_t/([A]_0/2)`
= `2.303/t_(1//2) log_10 2`
Substituting value of log102,
k = `2.303/t_(1//2) xx 0.3010`
∴ k = `0.693/t_(1//2)`
∴ `t_(1//2) = 0.693/k`
The half-life of a first-order reaction is independent of the initial reactant concentration.
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