Heats of combustion of benzene, carbon and hydrogen are `- 782, -94, -68 kcal mol^(-1)` respectively. Calculate heat of formation of benzene. 

Explanation

`C_(6)H_((s))+O_(2)(g)to 6CO_(2)(g)+3H_(2)O_((l)),DeltaH_(1)=-782k.cal`
`C_((s))+O_(2)(g)to CO_(2)(g),DeltaH_(2)=-94Kcal`
`H_(2)(g)+1/2O_(2)(g)to H_(2)O_((l)),DeltaH=-68Kcal `
The required equation is `6C_((s))+3H_(2)(g)to C_(6)H_(6)(l),DeltaH=?`
Heat of formation of benzene =`DeltaH=-DeltaH_(1)+6DeltaH_(2)+3DeltaH_(2)=+14Kcal`.