Chemistry · Chemical Kinetics

JEE Main 2025 — 22 January, Morning Shift — Question 46

$\mathrm{A} \rightarrow \mathrm{B}$ 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).

$\left[\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right.$ ]

Answer: 69

Numerical answer — enter this value.

$\quad t_{1 / 2}=\frac{0.693}{\mathrm{~K}}$

$\mathrm{K}=\mathrm{Ae}^{-\mathrm{Ea} / \mathrm{RT}}$

$=10^{20} \times \mathrm{e}^{-\frac{191.48 \times 10^{3}}{8.314 \times 1000}}$

$=10^{20} \times \mathrm{e}^{-23.031}=10^{20} \times-\mathrm{e}^{\ln 10 \times 10}$

$=\frac{10^{20}}{10^{10}}=10^{10} \mathrm{sec}$ .

$t_{1 / 2}=\frac{0.693}{10^{10}}=6.93 \times 10^{-11}$

$=69.3 \times 10^{-12} \mathrm{sec}$ .

Answer key and solution verified before publishing.

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