Question

The rate constant of a first order reaction increases five times when the temperature is raised from 350 K to 500 K. Calculate the activation energy of the reaction. (R = 8.314 JK⁻¹ mol⁻¹)

Answer 1

Given: R = 8.314 JK⁻¹ mol⁻¹

Initial temperature (T₁) = 350 K 

Final temperature (T₂) = 500 K

Rate constant ratio (k₂/k₁) = 5

Formula: `In(k_2/k_1) = E_a/R (1/T_1 - 1/T_2)`

`In (5) = E_a/(8.314) (1/350 - 1/500)`

`1.6094 = E_a/8.314 ((500 - 350)/(350 xx 500))`

`1.6094 = E_a/8.314 ((150)/(175000))`

`1.6094 = E_a/8.314 ((150)/(175000))`

`1.6094 = E_a/8.314 (0.000857)`

`E_a = (1.6094 xx 8.314)/0.000857`

= `13.38055/0.000857`

= 15613.2 J/mol

= `15613.2/1000`

= 15.61 kJ/mol