Calculations of Partial Molar Enthalpies from Experimental Data

Using the data in Fig. 8.1-1, determine the partial molar enthalpy of sulfuric acid and water at 50 mol % sulfuric acid and 65.6°C.

Question

Calculations of Partial Molar Enthalpies from Experimental Data

Using the data in Fig. 8.1-1, determine the partial molar enthalpy of sulfuric acid and water at 50 mol % sulfuric acid and 65.6°C.

Accepted Answer

First we must obtain values of enthalpy versus concentration at 65.6°C. The values read from this figure and converted to a molar basis are given below.

[latex] ext { wt } \% H _{2} SO _{4}[/latex]	[latex]\hat{H}\left(\frac{ kJ }{ kg } ight)[/latex]	[latex]\Delta_{\operatorname{mix}} \hat{H}\left(\frac{ kJ }{ kg } ight)[/latex]	[latex]x_{ H _{2} SO _{4}}[/latex]	[latex]\frac{ mol }{ kg }[/latex]	[latex]\Delta_{\operatorname{mix} \underline{H}}\left(\frac{ kJ }{ mol } ight)[/latex]
0	+278	0	0		0
20	+85	-155.8	0.0439	46.45	-3.354
40	-78	-281.6	0.1091	37.38	-7.533
60	-175	-341.4	0.216	28.32	-12.055
80	-153	-282.2	0.4235	19.26	-14.652
90	-60	-170.6	0.6231	14.73	-11.582
100	92	0	1.000		0

These data are fit reasonably well with the simple expression

[latex]\begin{aligned}\Delta_{ mix } \underline{H}\left(\frac{ kJ }{ mol } ight) &=x_{ H _{2} SO _{4}} x_{ H _{2} O }\left(-82.795+56.683 x_{ H _{2} SO _{4}} ight) \&=x_{ H _{2} SO _{4}}\left(1-x_{ H _{2} SO _{4}} ight)\left(-82.795+56.683 x_{ H _{2} SO _{4}} ight) \&=-82.795 x_{ H _{2} SO _{4}}+139.478 x_{ H _{2} SO _{4}}^{2}-56.683 x_{ H _{2} SO _{4}}^{3}\end{aligned}[/latex]

and

[latex]\frac{d \Delta_{\operatorname{mix}} \underline{H}}{d x_{ H _{2} SO _{4}}}=-82.795+278.965 x_{ H _{2} SO _{4}}-170.049 x_{ H _{2} SO _{4}}^{2}[/latex]

Therefore,

[latex]\left.\frac{d \Delta_{ mix } \underline{H}}{d x_{ H _{2} SO _{4}}} ight|_{x_{ H _{2} SO _{4}=0.5}}=14.17 \frac{ kJ }{ mol }[/latex]

and

[latex]\left.\frac{d \Delta_{ ext {mix }} \underline{H}}{d x_{ H _{2} O }} ight|_{x_{ H _{2} SO _{4}=0.5}}=-\left.\frac{d \Delta_{ mix } \underline{H}}{d x_{ H _{2} SO _{4}}} ight|_{x_{ H _{2} SO _{4}=0.5}}=-14.17 \frac{ kJ }{ mol }[/latex]

Also,

[latex]\Delta_{\operatorname{mix}} \underline{H}\left(x_{ H _{2} SO _{4}}=0.5 ight)=-13.61 \frac{ k J}{ mol }[/latex]

[latex]\underline{H}_{ H _{2} SO _{4}}=92 \frac{ kJ }{ kg } imes \frac{1 kg }{1000 g } imes \frac{98.708 g }{ mol }=9.02 \frac{ kJ }{ mol }[/latex]

and

[latex]\underline{H}_{ H _{2} O }=278 \frac{ kJ }{ kg } imes \frac{1 kg }{1000 g } imes \frac{18.015 g }{ mol }=5.01 \frac{ kJ }{ mol }[/latex]

Finally, from Eq. 8.6-9b, we have

[latex]\begin{aligned}\Delta_{ ext {mix }} \underline{H}-\left.x_{2} \frac{\partial\left(\Delta_{ ext {mix }} \underline{H} ight)}{\partial x_{2}} ight|_{T, P} &=\Delta_{ ext {mix }} \underline{H}+\left.x_{2} \frac{\partial\left(\Delta_{ ext {mix }} \underline{H} ight)}{\partial x_{1}} ight|_{T, P} \&=\left(\bar{H}_{1}-\underline{H}_{1} ight)\end{aligned}[/latex] (8.6-9b)

[latex]\begin{array}{l}\bar{H}_{ H _{2} SO _{4}}\left(x_{ H _{2} SO _{4}}=0.5, T=65.6^{\circ} C ight) \\quad=\underline{H}_{ H _{2} SO _{4}}\left(T=65.6^{\circ} C ight)+\Delta_{\operatorname{mix}} \underline{H}\left(x_{ H _{2} SO _{4}}=0.5 ight)+\left.x_{ H _{2} O } \frac{\partial \Delta_{ mix } \underline{H}}{\partial x_{ H _{2} SO _{4}}} ight|_{x_{ H _{2} SO _{4}}=0.5} \\quad=9.02-13.61+0.5(-14.17)=-11.68 \frac{ kJ }{ mol }\end{array}[/latex]

and

[latex]\begin{array}{l}\bar{H}_{ H _{2} O }\left(x_{ H _{2} SO _{4}}=0.5, T=65.6^{\circ} C ight) \=\underline{H}_{ H _{2} O }\left(T=65.6^{\circ} C ight)+\Delta_{ mix } \underline{H}\left(x_{ H _{2} SO _{4}}=0.5 ight)+\left.x_{ H _{2} SO _{4}} \frac{\partial \Delta_{ mix } \underline{H}}{\partial x_{ H _{2} O }} ight|_{x_{ H _{2} SO _{4}}=0.5} \=5.01-13.61+0.5(14.17)=-1.52 \frac{ kJ }{ mol }\end{array}[/latex]

Note that in this case the pure component and partial molar enthalpies differ considerably. Consequently, we say that this solution is quite nonideal, where, as we shall see in Chapter 9, an ideal solution is one in which some partial molar properties (in particular the enthalpy, internal energy, and volume) are equal to the pure component values. Further, here the solution is so nonideal that at the temperature chosen the pure component and partial molar enthalpies are even of different signs for both water and sulfuric acid. For later reference we note that, at [latex]x_{ H _{2} SO _{4}}=0.5[/latex], we have

[latex]\bar{H}_{ H _{2} SO _{4}}-\underline{H}_{ H _{2} SO _{4}}=-20.7 \frac{ kJ }{ mol } \quad ext { and } \quad \bar{H}_{ H _{2} O }-\underline{H}_{ H _{2} O }=-6.5 \frac{ kJ }{ mol }[/latex]

Question 8.6.1: Calculations of Partial Molar Enthalpies from Experimental D...

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