Correct Answer - C
`H_(2)C_(2)O(l) + (1)/(2)O_(2)(g) rarr H_(2)O(l) + 2O_(2)(g),`
`Deltan_(g)=3//2`
`DeltaU_(0) =-(0.312 xx8.75)/(1) xx 90 =-245.7 KJ//"mol"`
`DeltaH=DeltaU+ Deltan_(g)RT =- 245.7 KJ//"mol"`
`DeltaH=DeltaU + Deltan_(g)RT =-245.7 +(3)/(2) xx (8.314xx300)/(1000)=-241.947 KJ//"mol"`