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