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Question:

The rate constant for a first order reaction becomes six times when the temperature is raised from 350 K to 400 K. Calculate the activation energy for the reaction. [R=8.314 J K⁻¹mol⁻¹]

4.17 kJ mol⁻¹

4170 kJ mol⁻¹

41.7 kJ mol⁻¹

417.0 kJ mol⁻¹

Solution:

K = Ae⁻Ea/RT
K₂ = 6K₁
K₂/K₁ = e^(Ea/R) × (1/T₁ - 1/T₂)
6 = e^(Ea/R) × (1/350 - 1/400)
Ea = 41.7 kJ mol⁻¹