Question

A certain reaction is 50% complete in 20 minutes at 300 K and the same reaction is 50% complete in 5 minutes at 350 K. Calculate the activation energy if it is a first order reaction.

[R = 8.314 J K⁻¹ mol⁻¹, log 4 = 0.602]

Answer 1

Given: At 300 K, t₁ = 20 min

At 350 K, t₂ = 5 min

R = 8.314 J mol⁻¹ K⁻¹

log 4 = 0.602

Formula: `t_(1//2) = 0.693/k`

`k = 0.693/t_(1//2)`

∴ `k_1 = 0.693/20` = 0.03465

∴ `k_2 = 0.693/5` = 0.1386

`(k_2)/(k_1) = 0.1386/0.03465` = 4

 `log_(k_2//k_1) = E_a/(2.303 R) xx (1/T_1 - 1/T_2)`

`log 4 = E_a/(2.303 R) xx ((T_2 - T_1)/(T_1 xx T_2))`

`log 4 = E_a/(2.303 xx 8.314) xx( (350 - 300)/(300 xx 350))`

`0.602 = E_a/(19.147) xx( (50)/(105000))`

`0.602 = (E_a xx 50)/((19.147) (105000))`

`0.602 = (E_a xx 50)/(2010435)`

`E_a = (0.602 xx 2010435)/50`

`E_a = (1210281.87)/50`

= 24205.8 J mol⁻¹

= 24.205 kJ mol⁻¹