Enthalpy of solution from a thermodynamic cycle

Worked example

Use a thermodynamic cycle to find ΔsolH of MX₂(s), given ΔLH(dissociation) = 2414 kJ mol⁻¹, ΔhydH(M²⁺) = −1629 kJ mol⁻¹ and ΔhydH(X⁻) = −362 kJ mol⁻¹.

1. 1. Weighted sum of 1·ΔLH (dissociation) + 1·ΔhydH(M²⁺) + 2·ΔhydH(X⁻) → ΔsolH

   = 61.00 kJ mol⁻¹

Answer: ΔsolH = 61.0 kJ mol⁻¹

Method

MX₂(s) → M²⁺(g) + 2X⁻(g) → M²⁺(aq) + 2X⁻(aq). So ΔsolH = ΔLH(dissociation) + ΔhydH(M²⁺) + 2ΔhydH(X⁻). It is a small difference between large numbers, which is why solubility is hard to predict from these data alone.

Common mistakes

• MX₂ releases two X⁻ ions: include 2 × ΔhydH(X⁻).
• Dissolving first breaks the lattice apart (endothermic, +ΔLH dissociation), then hydrates the gaseous ions.

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