Question

The rate constant of a reaction is 1.2 × 10⁻³ s⁻¹ at 303 K and 2.1 × 10⁻³ s⁻¹ at 313 K. Calculate the energy of activation of the reaction. (R = 8.314 J K⁻¹ mol⁻¹)

Answer 1

Given:

k₁​ = 1.2 × 10⁻³ s⁻¹ at T₁ = 303 K

k₂ = 2.1 × 10⁻³ s⁻¹ at T₂ = 313 K

R = 8.314 J K⁻¹ mol⁻¹

Use the two-point form of the Arrhenius equation:

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

`ln((2.1 xx 10^-3)/(1.2 xx 10^-3)) = E_a/8.314 ((313 - 303)/(303 xx 313))`

`ln(1.75) = E_a/8.314 ((10)/(94839))`

`0.5606 = E_a/8.314 xx 1.0545 xx 10^-4`

⇒ `E_a = (0.5606 xx 8.314)/(1.0545 xx 10^-4)`

= `4.663/(1.0545 xx 10^-4)`

= 44,226 J/mol

E_a = 44.2 kJ mol⁻¹