JEE Main 2025 Chemistry question
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⁻¹ and the frequency factor is 10²⁰, the time required for 50%, molecules of A to become B is picoseconds (nearest integer). [R = 8.314 J K⁻¹ mol⁻¹]
Answer: 69
Solution
K = Ae^(-Ea/RT) = 10²⁰×e⁻²³·⁰³¹ = 10¹⁰ s⁻¹. t1/2 = 0.693/10¹⁰ = 69.3×10⁻¹² s ≈ 69 ps.