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