If a temperature increase from 21.0 ∘c to 35.0 ∘c triples the rate constant for a reaction, what is the value of the activation barrier for the reaction?

59.077 kJ/mol.

Explanation:

From Arrhenius law: K = Ae (-Ea/RT)

where, K is the rate constant of the reaction.

A is the Arrhenius factor.

Ea is the activation energy.

R is the general gas constant.

T is the temperature.

At different temperatures:

ln (k₂/k₁) = Ea/R [ (T₂-T₁) / (T₁T₂) ]

k₂ = 3k₁, Ea = ? J/mol, R = 8.314 J/mol. K, T₁ = 294.0 K, T₂ = 308.0 K.

ln (3k₁/k₁) = (Ea / 8.314 J/mol. K) [ (308.0 K - 294.0 K) / (294.0 K x 308.0 K) ]

∴ ln (3) = 1.859 x 10⁻⁵ Ea

∴ Ea = ln (3) / (1.859 x 10⁻⁵) = 59.077 kJ/mol.