2H2(g) + O2(g) → 2H2O(g)
\[\Delta H = \sum(\text{BE brutna bindn.}) - \sum (\text{BE bildade bindn.})\]
Brutna bindningar:
\[2 \cdot [\text{H-H]} + 1 \cdot [\text{O=O]} = (2 \cdot 436 + 498)\mathrm{kJ/mol} = 1370\mathrm{kJ/mol}\]
Bildade bindningar: 2 mol H2O innehåller 4 mol O–H-bindningar: 4 · [O–H] = 4 ⋅ 464 = 1856kJ/mol.
\[4 \cdot [\text{O-H]} = 4 \cdot 464\mathrm{kJ/mol} = 1856\mathrm{kJ/mol}\]
Totalt:
\[\Delta H = (1370 - 1856)\mathrm{kJ/mol} = -486\mathrm{kJ/mol}\]
