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JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics

\(\mathrm{A} \rightarrow \mathrm{~B}\)
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). \(\left[\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right]\)

  1. A 68
  2. B 69
  3. C 70
  4. D 71
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Answer & Solution

Correct Answer

(B) 69

Step-by-step Solution

Detailed explanation

\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…

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