Question

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?

Answer 1

Answer:

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.