Question

Solve

What is the energy of activation of a reaction whose rate constant doubles when the temperature changes from 303 K to 313 K?

Answer 1

Given:

Rate constants; k₂ = 2k₁,

Temperatures: T₁ = 303 K, T₂ = 313 K

To find:

Activation energy of the reaction (E_a)

Formula:

`"log"_10 "k"_2/"k"_1 = "E"_"a"/(2.303"R") (("T"_2" - T"_1)/("T"_2"T"_1))`

Calculation:

`"log" (2"k"_1)/"k"_1 = "E"_"a"/(2.303 xx 8.314 "J" "K"^-1 "mol"^-1) ((313 "K" - 303 "K")/(313 "K" xx 303 "K"))`

`"log" 2 = "E"_"a"/(2.303 xx 8.314  "J"  "mol"^-1) (10/(313 xx 303))`

0.3010 = `"E"_"a"/(19.147  "J"  "mol"^-1) xx 1.0544 xx 10^-4`

E_a = `(0.3010 xx 19.147)/(1.0544 xx 10^-4)` J mol^-1

E_a = 54659 J mol^-1 = 54.66 kJ mol^-1

The energy of activation of the reaction is 54.66 kJ mol^-1 .