Two substances $\mathrm{A}\left({\mathrm{t}}_{1/2}=5\min \right)\text{ and }\mathrm{B}\left({\mathrm{t}}_{1/2}=15\right.$min) follow first order kinetics, are taken in such a way that initially [A]=4[B].The time after which the concentration of both substances will be equal is :

For first order reaction, [R]_t=[R]_o e^-kt (or) C_t=C_oe^-kt

Consider, [A]=[B]

$$\begin{gathered} \mathrm{i}.\mathrm{e}. {\mathrm{C}}_{\mathrm{A}}{\mathrm{e}}^{-{\mathrm{k}}_{\mathrm{A}}\mathrm{t}}={\mathrm{C}}_{\mathrm{B}}{\mathrm{e}}^{-{\mathrm{k}}_{\mathrm{B}}\mathrm{t}} \\ \frac{{\mathrm{C}}_{\mathrm{A}}}{{\mathrm{C}}_{\mathrm{B}}}=\frac{{\mathrm{e}}^{-{\mathrm{k}}_{\mathrm{B}}\mathrm{t}}}{{\mathrm{e}}^{-{\mathrm{k}}_{\mathrm{A}}\mathrm{t}}}={\mathrm{e}}^{\left({\mathrm{k}}_{\mathrm{B}}-{\mathrm{k}}_{\mathrm{A}}\right)\mathrm{t}} \\ 4={\mathrm{e}}^{\left({\mathrm{k}}_{\mathrm{A}}-{\mathrm{k}}_{\mathrm{B}}\right)\mathrm{t}} (\mathrm{since}, [\mathrm{A}]=4[\mathrm{B}\left]\right) \\ \ln 4=\left({\mathrm{k}}_{\mathrm{A}}-{\mathrm{k}}_{\mathrm{B}}\right)\mathrm{t}=\left(\frac{\ln 2}{5}-\frac{\ln 2}{15}\right)\mathrm{t} \left(\mathrm{since}, \mathrm{k}=\frac{\ln 2}{ {\mathrm{t}}_{1/2}}\right) \\ \ln {\left(2\right)}^2=\ln 2\left(\frac{1}{5}-\frac{1}{15}\right)\mathrm{t} \\ 2\left(\ln 2\right)=\ln 2\left(\frac{1}{5}-\frac{1}{15}\right)\mathrm{t} \\ 2=\left(\frac{1}{5}-\frac{1}{15}\right)\mathrm{t} \\ \mathrm{t}=15\min . \end{gathered}$$

The time after which both the concentration will be equal is 15 minutes.