JEE Main2012ChemistryChemical KineticsActual
The activation energy for a reaction which doubles the rate when the temperature is raised from 298 ~K to 308 ~K is
Options
- A59.2 ~kJ ~mol ⁻¹
- B39.2 ~kJ ~mol ⁻¹
- C52.9 ~kJ ~mol ⁻¹
- D29.5 ~kJ ~mol ⁻¹
Correct answer
C. 52.9 ~kJ ~mol ⁻¹
Step-by-step solution
Activation energy can be calculated from the equation. aligned & K₂ K₁ = -E_a 2.303 R ( 1 T₂ - 1 T₁ ) & Given K₂ K₁ =2 T₂=308 ; T₁=298 & 2= -E_a 2.303 8.314 ( 1 308 - 1 298 ) & E_a=52.9 ~kJ ~mol ⁻¹ aligned