Enthalpy And Hess's Law

Question

The standard enthalpies of formation of CO₂(g) and H₂O(l) are −393.5 kJ mol⁻¹ and −285.8 kJ mol⁻¹ respectively. If the standard enthalpy of combustion of C₂H₂(g) (acetylene) is −1299.6 kJ mol⁻¹, then ΔH°_f of C₂H₂(g) is:

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1. **Write the Combustion Reaction**: C₂H₂(g) + 5/2 O₂(g) → 2CO₂(g) + H₂O(l).
2. **Apply the Relation**: $\Delta H_{comb} = \sum \Delta H_f(	ext{products}) - \sum \Delta H_f(	ext{reactants})$.
3. **Substitute Known Values**:
   $-1299.6 = [2(-393.5) + 1(-285.8)] - [\Delta H_f(C_2H_2) + 0]$
   $-1299.6 = [-787.0 - 285.8] - \Delta H_f(C_2H_2)$
   $-1299.6 = -1072.8 - \Delta H_f(C_2H_2)$.
4. **Solve**: $\Delta H_f(C_2H_2) = -1072.8 + 1299.6 = +226.8 \approx +226.7$ kJ mol⁻¹.
5. **Physical Meaning**: Acetylene has a positive enthalpy of formation, meaning it is thermodynamically less stable than its constituent elements — consistent with the highly exothermic nature of its combustion and the presence of a triple bond storing significant energy.

Enthalpy And Hess's Law

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