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

Hint: 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.