Using the appropriate bond energies, calculate the enthalpy change (ΔH) for the reaction: H_2(g) + Cl_2(g) → 2HCl(g). Given the following average bond energies: H–H = 436 kJ/mol, Cl–Cl = 242 kJ/mol, H–Cl = 431 kJ/mol.

• A. -92 kJ
• B. -184 kJ
• C. +184 kJ
• D. +92 kJ

Answer: (B)

Verified step by step guidance

1. Identify the bonds broken and bonds formed in the reaction. In the reaction H_2(g) + Cl_2(g) → 2HCl(g), the bonds broken are one H–H bond and one Cl–Cl bond, and the bonds formed are two H–Cl bonds.
2. Write down the bond energies given: H–H = 436 kJ/mol, Cl–Cl = 242 kJ/mol, and H–Cl = 431 kJ/mol.
3. Calculate the total energy required to break the bonds (bond breaking is endothermic, so energy is absorbed): \n\(\text{Energy to break bonds} = D_{\text{H–H}} + D_{\text{Cl–Cl}} = 436 + 242\) kJ/mol.
4. Calculate the total energy released when new bonds are formed (bond formation is exothermic, so energy is released): \n\(\text{Energy released forming bonds} = 2 \times D_{\text{H–Cl}} = 2 \times 431\) kJ/mol.
5. Determine the enthalpy change (\(\Delta H\)) for the reaction using the formula: \n\(\Delta H = \text{Energy to break bonds} - \text{Energy released forming bonds}\). A negative value indicates an exothermic reaction.