Enthalpy Change via Bond Energies Steps:

• Identify bonds broken: one H-H (436 kJ/mol) and one Cl₂ (242 kJ/mol); total energy input = 436 + 242 = 678 kJ/mol.
• Identify bonds formed: two H-Cl (431 kJ/mol each); total energy released = 2 × 431 = 862 kJ/mol.
• Calculate ΔH = energy to break bonds - energy from forming bonds = 678 - 862 = -184 kJ/mol.
• The negative value indicates an exothermic reaction.

Why B is correct:

• ΔH = -184 kJ/mol matches the calculation using the bond energy formula: sum of broken bonds minus sum of formed bonds.

Why the others are wrong:

• A: Ignores Cl-Cl bond and forming energies, underestimating the exothermic nature.
• C: Reverses the sign, treating the reaction as endothermic.
• D: Only considers H-H bond energy, ignoring all other bonds.

Final answer: B