JEE Main202522 Jan 2025Morning ShiftChemistryChemical KineticsActual
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 ]
Correct answer
0
Step-by-step solution
aligned & t _ 1 / 2 = 0.693 ~K & ~K = Ae ^ - Ea / RT & =10²⁰ e ^ - 191.48 10^3 8.314 1000 & =10²⁰ e ^ -23.031 =10²⁰ - e ^ 10 10 & = 10²⁰ 10¹⁰ =10¹⁰ sec . & t _ 1 / 2 = 0.693 10¹⁰ =6.93 10⁻¹¹ & =69.3 10⁻¹² sec . aligned