At `25^(@)C and 45^(@)C` , a first order reaction takes 30min and 10min respectively to reach 50% completion. Calculate the activation energy . If `DeltaH=-40"kJ.mol"^(-1)` , calculate the activation energy of backward reaction.

Explanation

`k_(2)=(0.693)/(10)"min"^(-1) and k_(1)=(0.693)/(30)"min"^(-1)`
`therefore (k_(2))/(k_(1))=3 therefore log3 = (E_(a))/(2.303R)((20)/(298xx318))`
`therefore E_(a)=43.28"kJ.mol"^(-1)`
Activation energy of backward reaction = Activation energy of the forward reaction ` +40kJ.mol^(-1)`
`=(43+28+40)"kJ.mol"^(-1)=83.28 "kJ.mol"^(-1)`