Question

The rate constant of a first order reaction becomes 6 times when the temperature is raised from 350 K to 410 K. Calculate the energy of activation. (Gas constant, R = 8.314 J K⁻¹ mol⁻¹)

Answer 1

Given:

`k_2/k_1 = 6`

T₁​ = 350 K, T₂ = 410 K

R = 8.314 J K⁻¹ mol⁻¹

We will use the Arrhenius equation in the two-point logarithmic form:

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

⇒ `ln(6) = E_a/8.314 ((410 - 350)/(350 xx 410))`

⇒ `1.7918 = E_a/8.314 (60/143500)`

⇒ `1.7918 = E_a/8.314 (4.18 xx 10^-4)`

`E_a = (1.7918 xx 8.314)/(4.18 xx 10^-4)`

= `14.9/(4.18 xx 10^-4)`

= 35,650 J/mol

= 35.5 kJ mol⁻¹