Question

Show that in a first order reaction, time required for completion of $75\%$ is twice of half-life of the reaction. $\left( {log2{\text{ }} = {\text{ }}0.3010} \right)$

Answer 1

Since, we know the expression for rate law for first order kinetics is given by $t = \dfrac{{2.303}}{k}\log \dfrac{a}{{a - x}}$ . So, for completion of half –life which is the amount of time taken by a radioactive material to decay to half of its original value, ${t_{\dfrac{1}{2}}} = \dfrac{{2.303}}{k}\log \dfrac{{100}}{{100 - 50}}$ and for completion of $75\%$ reaction, ${t_{75\% }} = \dfrac{{2.303}}{k}\log \dfrac{{100}}{{100 - 75}}$ . Now solving the ratio of time required for completion of $75\%$ to the half-life of the reaction, we can prove that in a first order reaction, time required for completion of $75\%$ is twice of half-life of the reaction.

Complete step by step answer:
The expression for rate law for first order kinetics is given by:
$t = \dfrac{{2.303}}{k}\log \dfrac{a}{{a - x}}$
where,
k = rate constant
t = age of sample
a = let initial amount of the reactant
a – x = amount left after decay process
a) for completion of half-life:
Half-life is the amount of time taken by a radioactive material to decay to half of its original value.
${t_{\dfrac{1}{2}}} = \dfrac{{2.303}}{k}\log \dfrac{{100}}{{100 - 50}} \\\ \Rightarrow {t_{\dfrac{1}{2}}} = \dfrac{{2.303}}{k}\log \dfrac{{100}}{{50}} \\\$
$\Rightarrow {t_{\dfrac{1}{2}}} = \dfrac{{2.303}}{k}\log 2$ … (1)
b) for completion of $75\%$ of reaction
${t_{75\% }} = \dfrac{{2.303}}{k}\log \dfrac{{100}}{{100 - 75}} \\\ \Rightarrow {t_{75\% }} = \dfrac{{2.303}}{k}\log \dfrac{{100}}{{25}} \\\ \Rightarrow {t_{75\% }} = \dfrac{{2.303}}{k}\log 4 \\\ \Rightarrow {t_{75\% }} = \dfrac{{2.303}}{k}\log {2^2} \\\$
$\Rightarrow {t_{75\% }} = \dfrac{{2.303}}{k}2\log 2$ … (2)
Therefore, From equation (1) and (2)
$\dfrac{{{t_{75\% }}}}{{{t_{\dfrac{1}{2}}}}} = \dfrac{{2\log 2}}{{\log 2}} \\\ \Rightarrow \dfrac{{{t_{75\% }}}}{{{t_{\dfrac{1}{2}}}}} = 2 \\\ \Rightarrow {t_{75\% }} = 2 \times {t_{\dfrac{1}{2}}} \\\$

Hence, it is proved that in a first order reaction, time required for completion of $75\%$ is twice of half-life of the reaction.

Note: A first-order reaction is a reaction that proceeds at a rate that depends linearly on only one reactant concentration and the half-life is independent of the initial concentration of reactant, which is a unique aspect to first-order reactions.