Question

Calculate activation energy for a reaction if its rate doubles when temperature is raised from 20°C to 35°C (R = 8.314 J K⁻¹ mol⁻¹).

Options

• 17.336 kJ

• 26.900 kJ

• 34.673 kJ

• 44.236 kJ

Answer 1

34.673 kJ

Explanation:

`ln  K_2/K_1 = E_a/R [1/T_1 - 1/T_2]`

`ln  (2K_1)/K_1 = E_a/8.314 [1/293 - 1/307]`

`E_a = ln 2 xx 8.314 [(293 xx 307)/(307 - 293)]`

E_a = 34.673 kJ