The heat of combustion of ethyl alcohol is `-300 kcal.` If the heats of formation of `CO_(2)(g)` and `H_(2)O(l)` are`-9.3` and `-68.5 kcal` respectively, calculate the heat of formation of ehtyl alcohol.

Answer

`(-94.1 kcal)`

Explanation

`C_(2)H_(6)O+3O_(2)(g)rarr2CO_(2)+3H_(2)O. " "DeltaH= -300 kcal`
`DeltaH_(comb)=2Delta H_(f,CO_(2))^(@)+3 DeltaH_(f,H_(2)O)^(@)-DeltaH_(f,C_(2)H_(6)O)`
`rArr DeltaH_(f,C_(2)H_(6)O)=2(-94.3)+3(-68.5)+300= -94.1 Kcal`