A arrow ~B The molecule A changes into its isomeric form B by...

\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]

Correct answer

Step-by-step 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}

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