Question
$A \rightarrow B$
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 $191.48 kJ mol ^{-1}$ and the frequency factor is $10^{20}$, the time required for $50 \%$, molecules of A to become B is _________ picoseconds (nearest integer).
$
\left[R=8.314 J K^{-1} mol^{-1}\right]
$

Answer

(69)
$
\begin{array}{l}Sol.\
t_{1 / 2}=\frac{0.693}{K} \\
K=Ae^{-EaRT} \\
=10^{20} \times e^{-\frac{199.48 \times 10^3}{8.314 \times 1000}} \\
=10^{20} \times e^{-23.031}=10^{20} \times-e^{\ln 10 \times 10} \\
=\frac{10^{20}}{10^{10}}=10^{10} sec \\
t_{1 / 2}=\frac{0.693}{10^{10}}=6.93 \times 10^{-11} \\
=69.3 \times 10^{-12} sec .
\end{array}
$

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