**Solve**

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

#### Solution

**Given: **

Rate constants; k_{2} = 2k_{1},

Temperatures: T_{1} = 303 K, T_{2} = 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} .