A substance A decomposes by two parallel first-order paths, A→ B with k1=2.0×10^-3 s^-1 and A→ C with k2=3.0×10^-3 s^-1. Starting with pure A, the…

For parallel first-order reactions, $\mathrm{A}$ disappears with the sum rate constant $k=k_1+k_2=5.0\times10^{-3}\ \mathrm{s^{-1}}$, so $[\mathrm{A}]=[\mathrm{A}]_0e^{-kt}$. The product formation rates are $d[\mathrm{B}]/dt=k_1[\mathrm{A}]$, so $[\mathrm{B}]=\dfrac{k_1}{k}[\mathrm{A}]_0\left(1-e^{-kt}\right)$; the branching ratio $k_1/k$ stays fixed throughout. Here $k_1/k=0.40$ and $kt=5.0\times10^{-3}\times200=1.00$, so $1-e^{-1}=0.6321$. Hence $[\mathrm{B}]/[\mathrm{A}]_0=0.40\times0.6321=0.2528$, i.e. $25.28\%$. A common error is to use $e^{-k_1t}$ alone, which neglects the competing path. Check: the fraction of A consumed is $1-e^{-1}=63.2\%$, of which $60\%$ ($37.9\%$) goes to C, and $36.8\%$ of A remains unreacted, so the three percentages sum to 100.