Consider the following parallel reactions being given by A (t_½ = 1.386 × 10² hours), each path being first order.
If the distribution of B in the product mixture is 50%, the partial half life of A for conversion into B is

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231 h

31 h

131 h

115.5 h

answer is A.

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Detailed Solution

$\begin{array}{l} \frac{2 k_{1}}{2 k_{1} + 3 k_{2}} = 0 .5 and \\ k_{1} + k_{2} = \frac{0 .693}{1 .386 \times 10^{2}} = 5 \times 10^{- 3} h^{- 1} \end{array}$

Solving $\frac{k_{1}}{k_{2}} = \frac{2}{3}$ and $k_{1} = 2 \times 10^{- 3} h^{- 1}$ and

$k_{2} = 3 \times 10^{- 3} h^{- 1}$

$t_{1 / 2 (A \to D)} = \frac{0 .693}{k_{2}} = \frac{0 .693}{3 \times 10^{- 3}} = 231 h$

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