20% first order reaction is completed in 50 minutes. Time required for the completion of 60% of the reaction is ....... min

$\mathrm{t}=\frac{2.303}{\mathrm{K}}\log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}$

At t_20% , a=100

             x=20

              t_{20% = 50min.}

  ${\mathrm{t}}_{20\%}=\frac{2.303}{\mathrm{K}}\log \left(\frac{100}{100-20}\right)$

${\mathrm{t}}_{20\%}=\frac{2.303}{\mathrm{K}}\log \left(\frac{5}{4}\right)$

At t_60%  a=100

             x= 60

  ${\mathrm{t}}_{60\%}=\frac{2.303}{\mathrm{K}}\log \left(\frac{100}{100-60}\right)$

${\mathrm{t}}_{60\%}=\frac{2.303}{\mathrm{K}}\log \left(\frac{5}{2}\right)$

$\frac{ {\mathrm{t}}_{60\%} = {\displaystyle \frac{2.303}{\mathrm{k}}} \log ({\displaystyle \frac{5}{2}} )}{ {\mathrm{t}}_{20\%}=\frac{2.303}{\mathrm{K}}\log \left(\frac{5}{4}\right)}$

$\frac{{\mathrm{t}}_{60\%}}{{\mathrm{t}}_{20\%}} =\frac{\log 5-\log 2}{\log 5-\log 4}$

${\mathrm{t}}_{60\%}=\left(\frac{0.397}{0.098}\right)50$

${\mathrm{t}}_{60\%}=205 \min .$