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

Explanation

The given data `C_6H_(6(s)) + 15//2O_(2(g)) to 6CO_(2(g)) + 3H_2O_((l)) , DeltaH_(1) = -782 K.cal`.
`C_((s)) + O_(2(g)) to CO_(2(g)) , DeltaH_2 = -94 kcal`
`H_(2(g)) + 1/2O_(2(g)) to H_2O_((l)) , DeltaH_(3) = -68 kcal`
The required equation is
`6C_((s)) + 3H_(2(g)) to C_6H_(6(l)) , DeltaH = ?`
Heat of formation of benzene =
`DeltaH = -DeltaH_1 + 6DeltaH_2 + 3DeltaH_2 = +14kcal`