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]

Solution

Chemistry ma’am · 4 steps

Get k from Arrhenius first, then 50% converted is the first-order half-life, 0.693/k.

1. Arrhenius in log form. Write k = A e^(-Eₐ/RT) with base-10 logs, so the numbers stay easy to handle. The energy barrier 191.48 kJ mol⁻¹ is Eₐ, and the frequency factor 10²⁰ is A.

   log k = log A - Eₐ/2.303RT = 20 - 191480/(2.303 × 8.314 × 1000)

2. Find k. The denominator is 2.303 × 8.314 × 1000 = 19147 J mol⁻¹, and 191480 / 19147 ≈ 10. So log k = 20 - 10 = 10, which gives k = 10¹⁰ s⁻¹.
3. Half-life. For a first-order reaction, the time for 50% conversion is t_(1/2) = 0.693/k = 0.693/10¹⁰ = 6.93 × 10⁻¹¹ s.
4. So. Since 1 ps = 10⁻¹² s, t_(1/2) = 69.3 ps, which rounds to 69.

The common slip: 100. Taking the time as 1/k instead of 0.693/k gives 100 ps, not 69.

The idea to keep. Whenever a question asks for the time for 50% of a first-order reactant to react, it is asking for t_(1/2) = 0.693/k. Find k first, then convert the units at the end.

Answer from the NTA final answer key. Solution by PrepWiser.