Chemistry · 2025
JEE Main · 22 January 2025, Shift 1 · Q73
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…
$$\mathrm{A} \rightarrow \mathrm{~B}
$$
The molecule A changes into its isomeric form B by following a first order kinetics at a temperature of $\displaystyle 1000$ K. If the energy barrier with respect to reactant energy for such isomeric transformation is $\displaystyle 191.48 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and the frequency factor is $\displaystyle 10^{20}$, the time required for $\displaystyle 50 \%$ molecules of $\displaystyle A$ to become B is $\displaystyle \_\_\_\_$ picoseconds (nearest integer). $\displaystyle \left[\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right]$
Official answer
From NTA’s final answer key for this paper.
69
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JEE Main 2025 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.