Question

The rate of a reaction triples when the temperature changes from 298 K to 318 K. Calculate the energy of activation of the reaction, assuming that it does not change with temperature.

(Given: R = 8.314 JK⁻¹ mol⁻¹, log 3 = 0.4771)

Answer 1

Given: R = 8.314 JK⁻¹ mol⁻¹

log 3 = 0.4771

T₁ = 298 K

T₂ = 318 K

`(K_2)/(K_1) = 3`

To find: E_a = ?

Formula: `log  (K_2)/(K_1) = E_a/2.303 R [1/(T_1) - 1/(T_2)]`

`log 3 = (E_a)/(2.303 xx 8.314)[1/298 - 1/318]`

`log 3 = (E_a)/(2.303 xx 8.314)[(298 - 318)/(298 xx 318)]`

`0.4771 = (E_a)/(2.303 xx 8.314)[10/(298 xx 318)]`

`E_a = (0.4771 xx 2.303 xx 8.314 xx 298 xx 318)/10`

`E_a = 865678.7/10`

E_a = 86567.87 J mol⁻¹

E_a = 86.567 KJ mol⁻¹