A first order reaction has a specific reaction rate of $${10^{ - 2}}{s^{ - 1}}.$$   How much time will it take for $$20$$ $$g$$ of the reactant to reduce to $$5$$ $$g?$$

Solution :
$$\eqalign{ & {\text{For a first order reaction,}} \cr & {\text{Rate constant }}\left( k \right) = \frac{{2.303}}{t}{\text{log}}\frac{a}{{a - x}} \cr & {\text{where, }}a = {\text{initial concentration}} \cr & a - x = {\text{concentration after time}}\,'t' \cr & t = {\text{time in sec}}{\text{.}} \cr & {\text{Given,}}\,a = 20\,g,\,a - x = 5\,g,\,k = {10^{ - 2}} \cr & \therefore \,\,t = \frac{{2.303}}{{{{10}^{ - 2}}}}{\text{log}}\,\frac{{20}}{5} \cr & \,\,\,\,\,\,\,\,\,\, = 138.6\,s \cr & {\text{Alternatively,}} \cr & {\text{Half - life for the first order reaction,}} \cr & \frac{{{t_{\frac{1}{2}}}}}{2} = \frac{{0.693}}{k} \cr & \,\,\,\,\,\,\,\, = \frac{{0.693}}{{{{10}^{ - 2}}}} \cr & \,\,\,\,\,\,\,\, = 69.3s \cr} $$
Two half-lives are required for the reduction of $$20$$ $$g$$ of reactant into $$5$$ $$g.$$
\[20\,g\xrightarrow{{{t}_{\frac{1}{2}}}}10\,g\xrightarrow{{{t}_{\frac{1}{2}}}}5\,g.\]
$$\therefore {\text{Total time}} = 2{t_{\frac{1}{2}}} = 2 \times 69.3 = 138.6\,s$$