A first order reaction is $\displaystyle 50$% complete in $\displaystyle 40$ minutes. Calculate the time required for the completion of $\displaystyle 90$% of reaction. [Given : $\displaystyle \log 2=0.3010, \log 10=1$ ]

\[\begin{aligned} & \mathrm{t}_{1 / 2}=0.693 / \mathrm{k} \\ & \mathrm{k}=0.693 / \mathrm{t}_{1 / 2}=0.693 / 40 \mathrm{~min}^{-1} \end{aligned} \] $\displaystyle 90$% completion \[\begin{aligned} \mathrm{t} & =2.303 / \mathrm{k} \log \left[\mathrm{R}_{0}\right] /[\mathrm{R}] \\ & =2.303 / 0.693 \times 40 \times \log 100 / 10 \\ & =2.303 / 0.693 \times 40=132.9 \mathrm{~min} \end{aligned} \]