Bond Enthalpy Calculation for Water Formation Steps:

• Balance the reaction: 2H₂ + O₂ → 2H₂O (bonds: break 2 H-H and 1 O=O; form 4 O-H).
• Calculate bonds broken energy: (2 × 436 kJ/mol) + 496 kJ/mol = 1368 kJ/mol.
• Calculate bonds formed energy: 4 × 463 kJ/mol = 1852 kJ/mol.
• Compute ΔH: 1368 kJ/mol - 1852 kJ/mol = -484 kJ/mol.

Why B is correct:

• Matches the bond enthalpy formula ΔH = Σ(bond energies broken) - Σ(bond energies formed), indicating exothermic reaction.

Why the others are wrong:

• A: Uses only 1 H-H bond broken (436 + 496 - 1852 = -920), ignoring stoichiometry for 2H₂O.
• C: Subtracts only 2 O-H bonds (1368 - 926 = 442), undercounting bonds formed in 2H₂O.
• D: Reverses sign of correct calculation, treating reaction as endothermic.

Final answer: B