Given the bond dissociation enthalpies (bde) below, what is the approximate ∆hºf for h2o(g)?

We can get ∆H for 2 H2 (g) + O2 (g) → 2 H2O (g) if the given bond dissociation enthalpies are 
    Bond     BDE, kJ mol-1      Bond      BDE, kJ mol-1
    H–H      432                      O–O      146
    O–H      467                       O=O      495

In the reactants, two H–H bonds require 2*432 kJ = 864 kJ and one O=O requires 1*495 kJ = 495 kJ

In the products, four O–H bonds give off 4*467 kJ = 1868 kJ

The enthalpy change of the reaction is
     ∆H = ∑ bond energy of reactants - ∑ bond energy of products
           = (864 kJ + 495 kJ) - 1868 kJ 
           = 1359 kJ - 1868 kJ 
           = -509 kJ for 2 moles of H2O 
Therefore, the approximate ∆H for H2O(g) is -254.5 kJ per mole