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
