Worked Example 2: Hydrogen + Chlorine → Hydrogen Chloride

🧮 Worked Example 2: Hydrogen + Chlorine → Hydrogen Chloride

Calculate ΔH for: H₂ + Cl₂ → 2HCl

Bond energies given:

• H-H bond = 436 kJ/mol
• Cl-Cl bond = 242 kJ/mol
• H-Cl bond = 431 kJ/mol

Step 1: Identify bonds broken (reactants)

• 1 × H-H bond = 1 × 436 = 436 kJ
• 1 × Cl-Cl bond = 1 × 242 = 242 kJ
• Total energy in: 436 + 242 = 678 kJ

Step 2: Identify bonds formed (products)

• 2HCl contains 2 × H-Cl bonds = 2 × 431 = 862 kJ
• Total energy out: 862 kJ

Step 3: Calculate ΔH

ΔH = 678 − 862 = −184 kJ/mol

Conclusion: ΔH is negative, so the reaction is exothermic. More energy is released forming H-Cl bonds than is needed to break the H-H and Cl-Cl bonds. This reaction can proceed spontaneously once initiated (e.g., by UV light or a spark).

Compare with the first example: Both are exothermic, but ΔH for H₂ combustion (−482 kJ/mol) is much larger than for H₂ + Cl₂ (−184 kJ/mol). This is because the O-H bonds formed in water (463 kJ/mol each) are much stronger than H-Cl bonds (431 kJ/mol).