Experimentally determining some enthalpies of formation is very difficult, if not impossible, while combustion enthalpies are very easy to determine experimentally in a bomb calorimeter. Use the following heats of combustion to find the enthalpy of formation of methanol.

CH3OH(l) + 3/2 O2(g) → CO2(g) + 2 H2O(l) ΔHcombustion = -726.4 kJ

C(graphite) + O2(g) → CO2(g) ΔHcombustion = -393.5 kJ

H2(g) + 1/2 O2(g)→ H2O(l) ΔHcombustion = -285.8 kJ

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

ΔHformation =

Question

Experimentally determining some enthalpies of formation is very difficult, if not impossible, while combustion enthalpies are very easy to determine experimentally in a bomb calorimeter. Use the following heats of combustion to find the enthalpy of formation of methanol.

CH3OH(l) + 3/2 O2(g) → CO2(g) + 2 H2O(l)	   ΔHcombustion = -726.4 kJ

C(graphite) + O2(g) → CO2(g)	   ΔHcombustion = -393.5 kJ

H2(g) + 1/2 O2(g)→ H2O(l)	   ΔHcombustion = -285.8 kJ

C(graphite) + 2 H2(g) + 1/2 O2(g) → CH3OH(l)    

ΔHformation =