A reaction proceeds by first order, $$75\% $$  of this reaction was completed in $$32\,\min .$$  The time required for $$50\% $$  completion is

Solution :
Given : 75% reaction gets completed in $$32\,\min $$
$$\eqalign{ & {\text{Thus,}}\,k = \frac{{2.303}}{t}{\text{log}}\frac{a}{{\left( {a - x} \right)}} \cr & = \frac{{2.303}}{{32}}{\text{log}}\frac{{100}}{{\left( {100 - 75} \right)}} \cr & = \frac{{2.303}}{{32}}{\text{log}}4 \cr} $$
  \[=0.0433\,{{\min }^{-1}}\]
Now we can use this value of $$k$$  to get the value of time required for 50% completion of reaction
$$\eqalign{ & t = \frac{{2.303}}{k}{\text{log}}\frac{a}{{\left( {a - x} \right)}} = \frac{{2.303}}{{0.0433}}{\text{log}}\frac{{100}}{{50}} \cr & = \frac{{2.303}}{{0.0433}}{\text{log}}\,2 = 16\,\min \cr} $$