Enthalpy Change via Bond Energies for H₂ + ½O₂ → H₂O(g)

Steps:

• Identify bonds broken: one H–H (436 kJ/mol) and half O=O (½ × 496 = 248 kJ/mol); total energy input = 684 kJ/mol.
• Identify bonds formed: two O–H (2 × 463 = 926 kJ/mol); total energy released = 926 kJ/mol.
• Calculate ΔH = energy to break bonds – energy from forming bonds = 684 – 926 = –242 kJ/mol.
• Confirm for one mole of H₂ reacting to form one mole of H₂O vapor.

Why C is correct:

• Matches the formula ΔH = Σ(bond energies broken) – Σ(bond energies formed), yielding exothermic –242 kJ/mol for water vapor formation.

Why the others are wrong:

• A: Positive value ignores exothermic nature; likely misapplies sign for endothermic.
• B: Equals H–H bond energy alone, omitting O=O and O–H contributions.
• D: Negative H–H energy alone, ignoring full reaction bonds.

Final answer: C