Study the parallel first order reactions:
$\begin{array}{l} A \overset{k_{1}}{⟶} 2 B \\ 2 A \overset{k_{2}}{⟶} 3 C \end{array}$
At any time $(t \neq 0)$ , the products contain 40% B and 60% C, by moles. If the overall rate constant for the reaction of A is $0.09 \min^{- 1}$ , then the value of $K_{1}$ is:

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$0.036 \min^{- 1}$

$0.054 \min^{- 1}$

$0.03 \min^{- 1}$

$0.06 \min^{- 1}$

answer is B.

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Detailed Solution

$A \overset{k_{1}}{⟶} 2 B; 2 A \overset{k_{2}}{⟶} 3 C a - x - y 2 x a - x - y \frac{3 y}{2} \begin{array}{l} - \frac{d [A]}{d t} = \frac{d x}{d t} + \frac{d y}{d t} = \frac{1}{2} \frac{d [B]}{d t} + \frac{2}{3} \frac{d [C]}{d t} \\ \Rightarrow - \frac{d [A]}{d t} = k_{1} [A] + 2 k_{2} [A] = (k_{1} + 2 k_{2}) [A] \\ \frac{[B]}{[C]} = \frac{4 x}{3 y} = \frac{40}{60} \\ \Rightarrow \frac{x}{y} = \frac{1}{2} \\ \Rightarrow \frac{d x}{d t} = \frac{1}{2} \frac{d y}{d t} \\ \Rightarrow k_{1} = k_{2} \\ \Rightarrow k_{1} + 2 k_{2} = 0.09 \\ \Rightarrow k_{1} = 0.03 \min^{- 1} \end{array}$