calculate sandard enthalpy of formation of CH3OH from:
CH3OH + 3/2O2 ---- CO2 + 2H2O delta r H = -726 Kj/Mol
C + O2 ---- CO2 dlta fH = -393.5 Kj/mol
H2 + 1/2O2 ----- H2O dlta fH = -286 Kj/Mol
to calculate the standard heat of formation of methanol your equations should be arranged like this
CO2 + 2H2O → CH3OH + 3/2O2 =ΔHf, 726 Kj/Mol (just reversing the equation.)
C + O2→ CO2 ΔHf, =-393.5 Kj/mol
2H2 + O2 →2H2O = ΔHf,-286 x2 = -572 Kj/Mol(multiplying the equation by two)
Now by summing up all 3 equations we get
C + 1/2O2 + 2H2→CH3OH = ( 726-393.5-572 = -239)KJmol-1