The rate of a particular reaction doubles when temperature changes from `27^(@)` C to `37^(@)C`. Calculate the energy of activation of such a reaction.

Explanation

Here, we are given that
When `T_(1)=27^(@)C=300" K, "k_(1)=k("say"). "When "T_(2)=37^(@)C=310" K",k_(2)=2k`
Substituting these values in the equation :
`log""(k_(2))/(k_(1))=(E_(a))/(2.303R)((T_(2)-T_(1))/(T_(1)T_(2))),`
we get, `log""(2k)/(k)=(E_(a))/(2.303xx8.314)xx(10)/(300xx310("J mol"^(-1)))" or "log2=(E_(a))/(2.303xx8.314)xx(10)/(300xx310("J mol"^(-1)))`
This on solving gives `E_(a)=53598.6" J mol"^(-1)=53.6" kJ mol"^(-1)`