A substance A decomposes by two parallel first-order paths, Axrightarrowk1B and Axrightarrowk2C, with k1=3k2. The half-life for the disappearance of…

For parallel first-order paths the disappearance of A follows the total rate constant $k=k_1+k_2$, and $t_{1/2}=\ln2/k$, so $k=0.0693\ \mathrm{min^{-1}}$. The products form in the ratio of their rate constants, so $[\mathrm{C}]=\dfrac{k_2}{k_1+k_2}[\mathrm{A}]_0\left(1-e^{-kt}\right)=0.25\,[\mathrm{A}]_0(1-e^{-kt})$. The maximum $[\mathrm{C}]$ is $0.25[\mathrm{A}]_0$, so $0.15$ is attainable. Setting $0.25(1-e^{-kt})=0.15$ gives $e^{-kt}=0.4$, hence $t=\ln2.5/k=0.9163/0.0693=13.2$ min. Using $k_2$ alone as the decay constant gives $9.4$ min; asking when $[\mathrm{A}]=0.85[\mathrm{A}]_0$ gives $2.3$ min; adding one half-life to the answer gives $23.2$ min. Note also that the branching ratio fixes $[\mathrm{B}]:[\mathrm{C}]=3:1$ throughout, so at that instant $[\mathrm{B}]=0.45[\mathrm{A}]_0$ and $[\mathrm{A}]=0.40[\mathrm{A}]_0$, which sum with $[\mathrm{C}]$ to unity.