Problem summaries

Calculate ΔfH°(CH3OH) using
Δ1H°= = –394 kJ mol–1
(1) C(s) + O2(g) → CO2(g)

Δ2H° = –286 kJ mol–1
(2) H2(g) + ½ O2(g) → H2O(l)

Δ3H°= –727 kJ mol–1
(3) CH3OH(l) + 3/2O2(g) → CO2(g) + 2H2O(l)


Solution 1 uses ΔrH° = ΣΔfH°(products) - ΣΔfH°(reactants)
The enthalpy changes Δ1H° and Δ2H° correspond to the enthalpies of formation for CO2 and H2O.
Δ3H° = ΔfH°(CO2) + 2 x ΔfH°(H2O) - ΔfH°(CH3OH) = Δ1H° + 2Δ2H° - ΔfH°(CH3OH)

Solution 2 uses combining of equations and their ΔH:
 (1) C(s) + O2(g) → CO2(g)ΔH° = Δ1H°
plus 2 × (4) 2H2(g) + O2(g) → 2H2O(l)ΔH° = 2 ×Δ2H°
plus(5) CO2(g) + 2H2O(l) → CH3OH(l) + 3/2OΔ5H° = – Δ3H°
equals (1) C(s) + 2H2(g) + ½O2(g) → CH3OH(l) ΔfH°(CH3OH) = Δ1H° + 2 ×Δ2H° – Δ3H°

ΔfH°(CH3OH) found by either method is -239 kJ mol-1.
Dr Sheila Woodgate
© The University of Auckland 2002