Question

A first order reaction is 50% complete in 30 minutes at 27°C and the same reaction is again 50% complete in 10 minutes at 47°C. Calculate the activation energy of the reaction.

(R = 8.314 JK⁻¹ mol⁻¹)

Answer 1

Given: t₁ ​= 30 min at T₁ = 27°C = 300 K

t₂ = 10 min at T₂ = 47°C = 320 K

For a first-order reaction, the half-life is:

t_1/2 = `0.693/k`

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

So,

k₁ = `0.693/30` = 0.0231 min⁻¹ and

k₂ = `0.693/10` = 0.0693 min⁻¹

By using the Arrhenius equation

`ln (k_2/k_1) = E_a/R (1/T_1 - 1/T_2)`

`ln (0.0693/0.0231) = E_a/8.314 (1/300 - 1/320)`

ln (3.00) = `E_a/8.314 ((320 - 300)/(300 xx 320))`

1.099 = `E_a/8.314 (20/96000)`

1.099 = `E_a/8.314 xx 2.083 xx 10^-4`

E_a = `(1.099 xx 8.314)/(2.083 xx 10^-4)`

= `9.13/(2.083 xx 10^-4)`

= 43.8 kJ mol⁻¹