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Q.

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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A→BThe 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]