A→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 10²⁰, the time required for 50%, molecules of A to become B is ______ picoseconds (nearest integer). [R = 8.314 J K^–1 mol^–1]

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answer is 69.

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Detailed Solution

$\begin{array}{l} t_{1 / 2} = \frac{0.693}{K} \\ K = A e^{- EaRT} \\ = 10^{20} \times e^{- \frac{19.488 \times 10^{3}}{8.314 \times 1000}} \\ = 10^{20} \times e^{- 23.031} = 10^{20} \times - e^{\ln 10 \times 10} \\ = \frac{10^{20}}{10^{10}} = 10^{10} \sec . \\ t_{1 / 2} = \frac{0.693}{10^{10}} = 6.93 \times 10^{- 11} \\ = 69.3 \times 10^{- 12} \sec . \end{array}$

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