Question

The following values for the first order reaction rate constants were obtained for a reaction:

Temp. (K)	k
298	3.46 × 10−5 s−1
308	13.50 × 10−5 s−1

Calculate the activation energy for the reaction.

Answer 1

Given:

T₁ = 298 K, k₁ = 3.46 × 10⁻⁵ s⁻¹

T₂ = 308 K, k₂ = 13.50 × 10⁻⁵ s⁻¹

R = 8.314 J mol⁻¹ K⁻¹

To calculate the activation energy (E_a) of a reaction using rate constants at two different temperatures, we use the Arrhenius equation in two-point form:

`ln(k_2/k_1) = E_a/R ((T_2 - T_1)/(T_1T_2))`

⇒ `ln((13.50 xx 10^-5)/(3.46 xx 10^-5)) = ln(13.50/3.46)`

= ln(3.9017)

≈ 1.360

Now apply in the equation:

`1.360 = E_a/8.314 ((308 - 298)/(298 - 308))`

= `E_a/8.314 (10/91784)`

= `E_a/8.314 * 1.089 xx 10^-4`

∴ `1.360 = (E_a xx 1.089 xx 10^-4)/8.314`

⇒ `E_a = (1.360 xx 8.314)/(1.089 xx 10^-4)`

= `11.312/(1.089 xx 10^-4)`

= 103,895 J/mol

= 103.9 kJ/mol