The heat of combustion of ethylene at `18^(@)C` and at constant volume is `-335.8` kcal when water is obtained in liquid state. Calculate the heat of combusion at constant pressure and at `18^@C`.

Explanation

The chemical equation for the combustion of `C_(2)H_(4)` is
`underset(1" mole")(C_(2)H_(4)(g))+underset("3 moles")(3O_(3)(g))=underset("2 moles")(2CO_(2)(g))+2H_(2)O(l),DeltaU=-335.8kcal`
No. of moles of reactants=(1+3)=4
No. of mols of products=2
So, `Deltan=(2-4)=-2`
Given `DeltaU=-335.8kcal,Deltan=-2,R=2xx10^(-3)kcal`
and `T=(18+273)=291K`
Applying `DeltaH=DeltaU+DeltanRT`
`=-335.8+(-2)(2xx10^(-3))(291)`
`=-336.964kcal`