The heat of combustion of ethylene at `18^(@)C` and at constant volume is `-330.0 kcal` when water is obtained in liquid state. Calculate the heat of combustion at constant pressure and at `18^(@)C`?

Explanation

The chemical equation for the combustion of `C_(2)H_(4)` is
`{:(C_(2)H_(4)(g)+,3O_(2)(g)=,2CO_(2)(g)+,2H_(2)O(l),DeltaU^(Theta) =- 330.8kcal,),(1mol,3mol,2mol,,):}`
Number of moles of reactants `= (1+3) = 4`
Number of moles pf products `= 2`
So, `Deltan = (2-4) =- 2`
Given `DeltaU^(Theta) =- 330.0 kcal, Deltan =- 2, R = 2 xx 10^(-3) kcal`
and `T = (18 +273) = 291 K`
Applying `DeltaH^(Theta) = DeltaU^(Theta) + DeltanRT`
`=- 330.0 + (-2) (2 xx 10^(-3)) (291)`
`=- 331.164 kcal`