Given the following heats of combustion.

Question

CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(l) ΔH°rxn = -726.4 kJ
C(graphite) + O2(g) CO2(g) ΔH°rxn = -393.5 kJ
H2(g) + 1/2 O2(g) H2O(l) ΔH°rxn = -285.8 kJ
Calculate the enthalpy of formation of methanol (CH3OH) from its elements.

C(graphite) + 2 H2(g) + 1/2 O2(g) CH3OH(l)

Answer 1

CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(l) ΔH°rxn = -726.4 kJ
C(graphite) + O2(g) CO2(g) ΔH°rxn = -393.5 kJ
H2(g) + 1/2 O2(g) H2O(l) ΔH°rxn = -285.8 kJ

Reverse eqn 1, add in eqn 2, add in 2x the reverse of eqn 3. That will give you the final equation you want. When you reverse an eqn you must change the sign of delta H. If you multiply an equation, multiply delta H by the same number.