Calculation of Infinite Dilution Partial Molar Enthalpies from Experimental Data
Compute the difference between the infinite dilution partial molar enthalpy and the pure component molar enthalpy for sulfuric acid and water at 65.6°C using the information in the previous illustration.
From the previous illustration
\left(\frac{\partial \Delta_{\operatorname{mix}} \underline{H}}{\partial x_{ H _{2} SO _{4}}}\right)_{T, P}=-82.795+278.965 x_{ H _{2} SO _{4}}-170.049 x_{ H _{2} SO _{4}}^{2}
Therefore,
\left(\frac{\partial \Delta_{\operatorname{mix}} \underline{H}}{\partial x_{ H _{2} SO _{4}}}\right)_{x_{ H _{2} SO _{4}}=1}=+26.11 \frac{ kJ }{ mol } \quad \text { and } \quad\left(\frac{\partial \Delta_{\operatorname{mix}} \underline{H}}{\partial x_{ H _{2} SO _{4}}}\right)_{x_{ H _{2} SO _{4}}=0}=-82.80 \frac{ kJ }{ mol }
so that
\bar{H}_{ H _{2} SO _{4}}\left(T=65.6^{\circ} C , x_{ H _{2} SO _{4}}=0\right)-\underline{H}_{ H _{2} SO _{4}}\left(T=65.6^{\circ} C \right)=-82.80 \frac{ kJ }{ mol }
in which case
\bar{H}_{ H _{2} SO _{4}}\left(T=65.6^{\circ} C , x_{ H _{2} SO _{4}}=0\right)=9.02-82.80=-73.80 \frac{ kJ }{ mol }
\bar{H}_{ H _{2} O }\left(T=65.6^{\circ} C , x_{ H _{2} O }=0\right)-\underline{H}_{ H _{2} O }\left(T=65.6^{\circ} C \right)=-26.11 \frac{ kJ }{ mol }
and
\bar{H}_{ H _{2} O }\left(T=65.6^{\circ} C , x_{ H _{2} O }=0\right)=5.01-26.11=-21.1 \frac{ kJ }{ mol }
Note that for the sulfuric acid + water system at T = 65.6°C the differences between the pure component and partial molar properties at infinite dilution are considerably greater than at the mole fraction of 0.5 in the previous illustration.