Example Question - 4

The enthalpy of formation of ethanol cannot be measured directly, because carbon, hydrogen and oxygen do not react together to form ethanol. Use the enthalpies of combustion below to calculate it.2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l)Δ_cH values: C(s) = -394 kJ mol^-1, H₂(g) = -286 kJ mol^-1, C₂H₅OH(l) = -1367 kJ mol^-1.Both sides of the equation burn to the same products, CO₂ and H₂O, so by Hess's law Δ_rH = ΣΔ_cH(reactants) - ΣΔ_cH(products). Oxygen has no enthalpy of combustion.Reactants: 2(-394) + 3(-286) = -788 + (-858) = -1646 kJ mol^-1. Products: -1367 kJ mol^-1.Δ_fH(ethanol) = -1646 - (-1367) = -279 kJ mol^-1. Note that the order of subtraction is the opposite of the formation method.