Question

If a temperature increase from 12.0 ∘c to 20.0 ∘c doubles the rate constant for a reaction, what is the value of the activation barrier for the reaction?

Answer 1

According to this formula:
㏑(K2/K1) = Ea / R / (1/T1 - 1/T2)

Ea = R ㏑(K2/K1) / (1/T1 - 1/T2)
When (K2/K1) the rate constant for the reaction = 2 (because it's doubled)
and we have R constant = 8.314 and T1= 12 + 273 = 285 kelvin T2 = 20 + 273=293 Kelvin
So by substitution:
∴ Ea (the activation barrier for the reaction) =( 8.314* ㏑(2) ) / (1/285 - 1/293)
= 60155 J Mol^-1 = 60 KJmol^-1