Consider the following parallel reactions being given by A \(=1.386 \times 10^{2}\) hours), each path being…

Consider the following parallel reactions being given by A \(=1.386 \times 10^{2}\) hours), each path being first order.

If the distribution of B in the Product mixture is 50%, the partial half life of A for converison into B is
  1. \(231 \mathrm{~h}\)
  2. \(131 \mathrm{~h}\)
  3. \(115.5 \mathrm{~h}\)
  4. \(31 \mathrm{~h}\)

Solution

$\begin{array}{l}
\frac{2 k_{1}}{2 k_{1}+3 k_{2}}=0.5 \text { and } \\
k_{1}+k_{2}=\frac{0.693}{1.386 \times 10^{2}}=5 \times 10^{-3} h^{-1}
\end{array}$
Solving $\frac{k_{1}}{k_{2}}=\frac{2}{3}$ and $k_{1}=2 \times 10^{-3} h^{-1}$ and
$\begin{array}{l}
k_{2}=3 \times 10^{-3} h^{-1} \\
t_{1 / 2(A ightarrow B)}=\frac{0.693}{k_{2}}=\frac{0.693}{3 \times 10^{-3}}=231 h
\end{array}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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