Given that the temperature coefficient for the saponification of ethylacetate by `NaOH` is `1.75`. Calculate the activation energy.

Explanation

`(k_(2))/(k_(1))= 1.75`, `T_(1)= 298 k, T_(2)= 308K`
(Since temperature coefficient is the ratio of rate cosntants at `35^(@)C and 25^(@)C)`
`2.303 log.(k_(2))/(k_(1))=(E_(a))/(R )[(T_(2)-T_(1))/(T_(1)T_(2))]`
`2.303 log 1.75=(E_(a))/(1.987)xx(308-298)/(308xx298)`
`E_(a)=(2.303xx308xx298xx1.987)/(10) log 1.75 cal mol^(-1)`
`= 10.207 kcal mol^(-1)`
`10.207 kcal mol^(-1)`