All question types · Thermodynamics
An unknown bond enthalpy from ΔH
A-Level (AQA) · unreviewed
H₂ + Cl₂ → 2HCl with ΔH and two of the three bond enthalpies known.
Worked example
Hydrogen and chlorine react: H₂(g) + Cl₂(g) → 2HCl(g), ΔrH = −183 kJ mol⁻¹. Use E(H–H) = 436 kJ mol⁻¹ and E(Cl–Cl) = 243 kJ mol⁻¹ to find E(H–Cl).
1. Weighted sum of 0.5·H–H + 0.5·Cl–Cl − 0.5·ΔrH → H–Cl
= 431.0 kJ mol⁻¹
Answer: E(H–Cl) = 431 kJ mol⁻¹
Method
ΔH = [E(H–H) + E(Cl–Cl)] − 2E(H–Cl), so E(H–Cl) = [E(H–H) + E(Cl–Cl) − ΔH]/2.
Common mistakes
- Two H–Cl bonds form, so the total for bonds made must be shared between two bonds.
- ΔH = broken − made, so made = broken − ΔH. With ΔH negative, the bonds made total more than the bonds broken.
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