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.

Explanation

Given that,
(i) `C_(10)H_(8)(s)+12O_(2)(g) to 10CO_(2)(g)+4H_(2)O(l),DeltaH=-123.25"kcal"`
(ii) `C(s)+O_(2)(g) to CO_(2)(g),DeltaH=-97.0"kcal"`
(iii) `H_(2)(g)+(1)/(2)O_(2)(g) to H_(2)O(l),DeltaH=-68.4"kcal"`
We have to calculate `DeltaH` of the equation,
`10C(s)+4H_(2)(g) to C_(10)H_(8)(s),DeltaH=?`
Applying the inspection method,
[-Eqn.(i)+10xxEqn. (ii)+4xxEqn.(iii)], we get,
`-C_(10)H_(8)(s)-12O_(2)(g)+10C(s)+10O_(2)(g)+4H_(2)(g)+2O_(2)(g) to -10CO_(2)(g)-4H_(2)O(l)+10CO_(2)(g)+4H_(2)O(l),DeltaH=-(-123.25)+10xx(-97.0)+4xx(-68.4)`
or `10C(s)+4H_(2)(g) to C_(10)H_(8)(s),DeltaH=-1120.35"kcal"`