StreakPeaked· Practice

ExamsJEE MainChemistry

For the decomposition COCl2(g) -> CO(g) + Cl2(g) with delta-H = 19 kcal/mol, the rate constant follows ln k = 15 - 5000/T. Calculate the activation energy (in kcal/mol) for the reaction.

  1. 9.94
  2. 5.00
  3. 19.0
  4. 10.0

Correct answer: 9.94

Solution

Matching ln k = ln A - Ea/(RT) gives Ea/R = 5000, so Ea = 5000 x R = 5000 x 1.987 cal/mol = 9935 cal/mol approx 9.94 kcal/mol.

Related JEE Main Chemistry questions

⚔️ Practice JEE Main Chemistry free + battle 1v1 →