The half-life for the thermal decomposition of acetone is 80 s and is independent of initial concentration of acetone. The time required for the reaction to go to 80% composition is (Given: log 2 = 0.30)

The reaction follows first-order kinetics. Hence $k=0.693/t_{1/2}=0.693/\left(80\mathrm{s}\right)$ For the reaction to be 80%, $[\mathrm{A}{]}_t/[\mathrm{A}{]}_0=0.2$ 

Hence

$$\begin{gathered} t=-\frac{2.303\log \left\{[\mathrm{A}{]}_t/[\mathrm{A}{]}_0\right\}}{k} \\ t=-\frac{\left(2.303\right)\left(\log 2\times {10}^{-1}\right)}{(0.693/80\mathrm{s})} \\ t=-\frac{\left(2.303\right)(0.301-1)}{(0.693/80\mathrm{s})} \\ t=186.1\mathrm{s} \end{gathered}$$