If a temperature increase from 10.0 ∘C to 21.0 ∘C doubles the rate constant for a reaction, what is the value of the activation barrier for the reaction

According to this formula:

㏑ (K2/K1) = Ea/R (1/T1 - 1/T2)

when K is the rate constant

Ea is the activation energy

R is the universal gas constant

and T is the temperature K

when K is doubled so K2: K1 = 2:1 & R = 8.314 J. K^-1. mol^-1

and T1 = 10 + 273 = 283 k & T2 = 21 + 273 = 294 k

So by substitution:

㏑2 = (Ea / 8.314) (1/283 - 1/294)

∴ Ea = 43588.9 J/mol = 43.6 KJ/mol