Chemistry · Ch 3 — Thermodynamics

Enthalpy of Dilution

3.5(f)

Enthalpy of Dilution

The enthalpy of solution is the enthalpy change associated with the addition of a specified amount of solute to a specified amount of solvent at constant temperature and pressure. This argument can be applied to any solvent with slight modification.

Example — HCl dissolving in water:

The enthalpy change for dissolving one mole of gaseous hydrogen chloride in 10 mol of water:

\text{HCl}(g) + 10\ \text{aq.} \rightarrow \text{HCl} \cdot 10\ \text{aq.}; \quad \Delta H = -69.01\ \text{kJ mol}^{-1}

Consider the following set of enthalpy changes:

Equation	Process	$\Delta H$ (kJ mol⁻¹)
S-1	$\text{HCl}(g) + 25\ \text{aq.} \rightarrow \text{HCl} \cdot 25\ \text{aq.}$	–72.03
S-2	$\text{HCl}(g) + 40\ \text{aq.} \rightarrow \text{HCl} \cdot 40\ \text{aq.}$	–72.79
S-3	$\text{HCl}(g) + \infty\ \text{aq.} \rightarrow \text{HCl} \cdot \infty\ \text{aq.}$	–74.85

The values of $\Delta H$ show a general dependence of the enthalpy of solution on the amount of solvent. As more and more solvent is used, the enthalpy of solution approaches a limiting value — the value in infinitely dilute solution.

Enthalpy of dilution: If we subtract equation S-1 from equation S-2:

\text{HCl} \cdot 25\ \text{aq.} + 15\ \text{aq.} \rightarrow \text{HCl} \cdot 40\ \text{aq.}

$\Delta H = [-72.79 - (-72.03)]\ \text{kJ mol}^{-1} = -0.76\ \text{kJ mol}^{-1}$ …

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