Bond Enthalpy

Question

The bond enthalpies of H–H, Cl–Cl, and H–Cl bonds are 436, 242, and 431 kJ mol⁻¹ respectively. The enthalpy change for the reaction H₂(g) + Cl₂(g) → 2HCl(g) is:

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1. **Bond Enthalpy Method**: $\Delta H = \sum(	ext{B.E. of bonds broken}) - \sum(	ext{B.E. of bonds formed})$.
2. **Bonds Broken** (reactant side):
   - 1 × H–H = 436 kJ
   - 1 × Cl–Cl = 242 kJ
   - Total = 678 kJ.
3. **Bonds Formed** (product side):
   - 2 × H–Cl = 2 × 431 = 862 kJ.
4. **Calculate**: $\Delta H = 678 - 862 = -184$ kJ mol⁻¹.
5. **Interpretation**: The reaction is exothermic because more energy is released during the formation of two strong H–Cl bonds than is consumed in breaking the weaker Cl–Cl and H–H bonds.
6. **Cross-check**: The known experimental value of ΔH for this reaction is approximately −185 kJ mol⁻¹, consistent with our calculation.

Bond Enthalpy

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