Calculate the activation energy of a reaction if the rate of the reaction is doubled while the temperature of the reaction system is increased from `27^(@)C" to " 37^(@)C`

Explanation

We know , `log.(k_(2))/(k_(1))=(E_(a))/(2.303R)((T_(2)-T_(1))/(T_(1)-T_(2)))`
Given, `T_(1)=(273+27)=300k,T_(2)(273+37)=310K and k_(2)=2k_(1)`
`thereforelog.(2k_(1))/(k_(1))=(E_(a))/(2.303xx8.314)((310-300)/(300xx310))`
or,` E_(a)=(2.303xx8.314xx300xx310)/(10)xxlog2`
`=53.6xx10^(3)"J.mol"^(-1)=53.6"kJ.mol"^(-1)`