A substance $$A$$  decomposes by a first order reaction starting initially with $$\left[ A \right] = 2.00\,m$$    and after $$200\,\min ,\left[ A \right]$$   becomes $$0.15\,m.$$  For this reaction $${t_{\frac{1}{2}}}$$ is

Solution :
\[\begin{align} & {{\left[ A \right]}_{0}}=2.0\,m,\,\left[ A \right]=0.15\,m,\,t=200\,\min \\ & \text{For first order reaction} \\ & {{A}_{0}}=\text{Initial concentration} \\ & A=\text{Final concentration} \\ & \text{Rate constant,} \\ & k=\frac{2.303}{t}\text{log}\frac{{{\left[ A \right]}_{0}}}{\left[ A \right]} \\ & \,\,\,\,\,=\frac{2.303}{200}\text{log}\frac{2.0}{0.15} \\ & \,\,\,\,\,=\frac{2.303}{200}\left( \text{log}\,200-\text{log}\,15 \right) \\ & \,\,\,\,\,=\frac{2.303}{200}\times \left( 2.3010-1.1761 \right) \\ & \,\,\,\,\,=\frac{2.303\times 1.1249}{200} \\ & \,\,\,\,\,=0.01295\,{{\min }^{-1}} \\ & \text{Now, half life,} \\ & {{t}_{\frac{1}{2}}}=\frac{0.6932}{k} \\ & \,\,\,\,\,\,\,=\frac{0.6932}{0.01295} \\ & \,\,\,\,\,\,\,=53.50\,\min \\ \end{align}\]