Question

The half-lives of a first-order reaction are 1.19s at 313 K and 15.45s at 293 K. Calculate the energy of activation.

Answer 1

`log_10  K_2/K_1 = E_1/(2.303 R)[(T_2 - T_1)/(T_1T_2)]`

`(t_(1/2))_1 = 0.693/K_1 and (t_(1/2))_2 = 0.693/K_2   ∴ ((t_(1/2))_1)/((t_(1/2))_2) = K_2`

`∴ log_10 = ((t_(1/2))_1)/((t_(1/2))_2) = E_a/(2.303 R)[(T_2 - T_1)/(T_1T_2)]`

Given: `(t_(1/2))_1 = 15.45  "s at"  T_1 = 293 K and (t_(1/2))_1 = 1.19   "s at"  T_2 = 313 K`

`∴ log_10  15.45/1.19 = E_a/(2.303 xx 8.314)[(313 - 293)/(313 xx 293)]`

`log_10  (12.983) = E_a/19.147 xx 20/(313 xx 293)`

`E_a = (1.1134 xx 19.147 xx 313 xx 293)/20 = 97752  "J mol"^-1 = 97.75  "kJ mol"^-1`