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). \(\left[\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right]\)
- A 68
- B 69
- C 70
- D 71
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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