Question: What is the change in enthalpy (ΔH) for the reaction: 2H2(g) + O2(g) → 2H2O(g) if the bond enthalpies are: H-H = 436 kJ/mol, O=O = 498 kJ/mol, H-O = 463 kJ/mol?
Options:
Correct Answer: −572 kJ
Solution:
ΔH = Σ(bond enthalpies of reactants) - Σ(bond enthalpies of products) = [2(436) + 498] - [4(463)] = −572 kJ.