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Chemical Kinetics and Nuclear Chemistry question

2025 · 22 Jan · Shift 1 · Q24
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Chemical Kinetics and Nuclear Chemistry question

2025 · 22 Jan · Shift 1 · Q24

JEE MainChemistryChemical Kinetics and Nuclear ChemistryNumerical+4 / −1
A→B\mathrm{A \rightarrow B}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−1191.48 \mathrm{~kJ} \mathrm{~mol}^{-1}191.48 kJ mol−1 and the frequency factor is 102010^{20}1020, the time required for 50%50 \%50% molecules of A to become B is ‾\underline{\hspace{2cm}}​ picoseconds (nearest integer). [R=8.314 J K−1 mol−1]\left[\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right][R=8.314 J K−1 mol−1]
Numerical answer
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