$\mathrm{A} \rightarrow \mathrm{~B}$ The molecule A changes into its isomeric form B by following a first…

$\mathrm{A} \rightarrow \mathrm{~B}$
The molecule A changes into its isomeric form B by following a first order kinetics at a temperature of 1000 K. If the energy barrier with respect to reactant energy for such isomeric transformation is $191.48 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and the frequency factor is $10^{20}$, the time required for $50 \%$ molecules of A to become $B$ is _________ picoseconds (nearest integer). $\left[\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right]$

Solution

$\begin{aligned} & \mathrm{t}_{1 / 2}=\frac{0.693}{\mathrm{~K}} \\ & \mathrm{~K}=\mathrm{Ae}^{-\mathrm{Ea} / \mathrm{RT}} \\ & =10^{20} \times \mathrm{e}^{-\frac{191.48 \times 10^3}{8.314 \times 1000}} \\ & =10^{20} \times \mathrm{e}^{-23.031}=10^{20} \times-\mathrm{e}^{\ln 10 \times 10} \\ & =\frac{10^{20}}{10^{10}}=10^{10} \mathrm{sec}. \\ & \mathrm{t}_{1 / 2}=\frac{0.693}{10^{10}}=6.93 \times 10^{-11} \\ & \quad=69.3 \times 10^{-12} \mathrm{sec}.\end{aligned}$

Asked in: JEE Main 2025 (22 Jan Shift 1)

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