Q.Prove that in case of first order reaction $t_{99.9\%} = 10\,t_{1/2}$.

Writing both the 99.9%-completion time and the half-life using the first-order integrated rate law, in terms of the same rate constant $k$, and dividing one by the other shows their ratio is almost exactly 10.

Integrated rate law for a first order reaction: $k = \dfrac{2.303}{t}\log\dfrac{[A]_0}{[A]}\quad\Rightarrow\quad t = \dfrac{2.303}{k}\log\dfrac{[A]_0}{[A]}$

Time for 99.9% completion ($t_{99.9\%}$): when the reaction is 99.9% complete, only $0.1\%$ of the reactant remains, i.e. $[A]=0.001[A]_0$, so $[A]_0/[A] = 1000$: $t_{99.9\%} = \dfrac{2.303}{k}\log(1000) = \dfrac{2.303}{k}\times3 = \dfrac{6.909}{k}$

Half-life ($t_{1/2}$): at 50% completion, $[A]_0/[A]=2$: $t_{1/2} = \dfrac{2.303}{k}\log(2) = \dfrac{2.303\times0.301}{k} = \dfrac{0.693}{k}$

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