Q.14. Calculate the standard enthalpy of formation of CH3OHf from the following data:

CH3OHl+32O2gCO2g+2H2Ol;rHθ=-726kJmol-1
Cg+O2gCO2g;cHθ=-393kJmol-1
H2g+12O2gH2Of;fHθ=-286kJmol-1.

NEETprep Answer:
The reaction takes place during the formation of CH3OHl can be written as:
Cs+2H2Og+12O2gCH3OHl1
The reaction (1) can be obtained from the given reactions by following the algebraic calculations as:
Equation (ii) + 2 × equation (iii) – equation (i)
fHθCH3OHl=cHθ+2fHθH2Ol-rHθ
=-393kJmol-1+2-286kJmol-1--726kJmol-1
=-393-572+726kJmol-1
fHθCH3OHl=-239kJmol-1