Question

It is generally observed that the rate of a chemical reaction becomes double with every 10°C rise in temperature. If the generalisation holds true for a reaction in the temperature range of 298 K to 308 K, what would be the value of activation energy (E_a) for the reaction?

In general, it is observed that the rate of a chemical reaction becomes double with every 10°C rise in temperature. If this generalisation holds correct for a reaction, calculate the value of activation energy when temperature changes from 298 K to 308 K. (R = 8.314 J K⁻¹ mol⁻¹)

Answer 1

Given: T₁ = 298 K

T₂ = 308 K

`K_2/K_1` = 2

R = 8.314 J K⁻¹ mol⁻¹

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

`log 2 = E_a/(2.303 xx 8.314) [1/298 - 1/308]`

`0.3010 = E_a/19.147143[(308 - 298)/(298 xx 308)]`

`0.3010 = E_a/19.147143[(10)/(298 xx 308)]`

`0.3010 = E_a/19.147143[(10)/(91784)]`

`E_a = (0.3010 xx 19.147143 xx 91784)/10`

`E_a = (528977.8)/10`

E_a = 52897.78 J mol⁻¹

E_a = 52.89778 kJ mol⁻¹ ≈ 52.898 kJ mol⁻¹