Question

A first-order reaction is 50% complete in 30 minutes at 300 K and in 10 minutes at 320 K. Calculate activation energy (E_a) for the reaction. [R = 8.314 J K⁻¹ mol⁻¹]

[Given: log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021]

A first order reaction is 50% completed in 30 minutes at 300 K and in 10 minutes at 320 K. Calculate the activation energy of the reaction. (R = 8.314 J K⁻¹ mol⁻¹)

Answer 1

Given: T_1/2 = 30 min, T₁ = 300 K

T_1/2 = 10 min, T₂ = 320 K

Formula: `log  k_2/k_1 = E_a/(2.303 R) [1/T_1 - 1/T_2]`

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

k₁ at 300 K = `0.693/30` min⁻¹

k₂ at 320 K = `0.693/10` min⁻¹

`log (k_2/k_1) = E_a/(2.303 R) [1/T_1 - 1/T_2]`

`log ((0.693/10)/(0.693/30)) = E_a/(2.303 R) [1/300 - 1/320]`

log (3) = `E_a/(2.303 R) [20/(300 xx 320)]`

0.4771 = `E_a/(2.303 xx 8.314) [20/(96000)]`

0.4771 = `E_a/(2.303 xx 8.314) [20/(96000)]`

E_a = `(0.4771 xx 2.303 xx 8.314 xx 96000)/20`

= `876969.73/20`

= 43848.48 J/mol

E_a = 43.8 kJ mol⁻¹

Hence, the activation energy of the reaction is 43.8 kJ⁻¹.