Rate of the chemical reaction doubles for an increase of 10K from 298K . Calculate `E_(a)`.

Explanation

Given, at `T_(1) = 298K, k_(1) = k ` (assume)
`therefore " at " T_(2)=308K,k_(2)=2k and E_(a)=?`
`therefore` Putting these values into the reaction `log.(k_(2))/(k_(1))=(E_(a))/(2.303R)`
we have , `[(T_(2)-T_(1))/(T_(1)T_(2))]`
or, `E_(a)=(2.303xx8.314xx308xx298xx0.3010)/(10)`
`=25.898xx10^(3)"J.mol"^(-1)=52.898"kJ.mol"^(-1)`