The enthalpy of a binary liquid system of species 1 and 2 at fixed T and P is represented by the equation: H = 400 x1 + 600 x2 + x1 x2 ( 40 x1 + 20 x2 ) where H is in J·mol^–1. Determine expressions for H1 and H2 as functions of x1, numerical values for the pure-species enthalpies H1 and H2,and

Question

The enthalpy of a binary liquid system of species 1 and 2 at fixed T and P is represented by the equation:

[latex] H=400 x_1+600 x_2+x_1 x_2\left(40 x_1+20 x_2 ight) [/latex]

where H is in [latex]J·mol^{–1}[/latex]. Determine expressions for [latex] \bar{H}_1 ext { and } \bar{H}_2 [/latex] as functions of [latex]x_1[/latex], numerical values for the pure-species enthalpies [latex]H_1 and H_2[/latex], and numerical values for the partial enthalpies at infinite dilution [latex] \bar{H}_1^{\infty} ext { and } \bar{H}_2^{\infty} ext {. } [/latex]

Accepted Answer

Replacing [latex]x_2 by 1 – x_1[/latex] in the given equation for H and simplifying gives:

[latex] H=600-180 x_1-20 x_1^3 [/latex] (A)

[latex] \frac{d H}{d x_1}=-180-60 x_1^2 [/latex]

By Eq. (10.15),

[latex] \bar{M}_1=M+x_2 \frac{d M}{d x_1} [/latex] (10.15)

[latex] \bar{H}_1=H+x_2 \frac{d H}{d x_1} [/latex]

Then,

[latex] \bar{H}_1=600-180 x_1-20 x_1^3-180 x_2-60 x_1^2 x_2 [/latex]

Replacing [latex]x_2 with 1 – x_1[/latex] and simplifying:

[latex] \bar{H}_1=420-60 x_1^2+40 x_1^3 [/latex] (B)

By Eq. (10.16),

[latex] \bar{M}_2=M-x_1 \frac{d M}{d x_1} [/latex] (10.16)

[latex] \bar{H}_2=H-x_1 \frac{d H}{d x_1}=600-180 x_1-20 x_1^3+180 x_1+60 x_1^3 [/latex]

or

[latex] \bar{H}_2=600+40 x_1^3 [/latex] (C)

One can equally well start with the given equation for H. Because [latex]dH/dx_1[/latex] is a total derivative, [latex] x_2[/latex] is not a constant. Also, [latex]x_2 = 1 – x_1[/latex] ; therefore [latex] dx_2/dx_1 = –1.[/latex] Differentiation of the given equation for H therefore yields:

[latex] \frac{d H}{d x_1}=400-600+x_1 x_2(40-20)+\left(40 x_1+20 x_2 ight)\left(-x_1+x_2 ight) [/latex]

Replacing [latex]x_2 with 1 – x_1[/latex] reproduces the expression previously obtained.

A numerical value for [latex]H_1[/latex] results by substitution of [latex]x_1 = 1[/latex] in either Eq. (A) or (B). Both equations yield [latex]H_1 = 400 J·mol^{–1}[/latex]. Similarly, [latex]H_2[/latex] is found from either Eq. (A) or (C) when [latex]x_1 = 0[/latex]. The result is [latex]H_2 = 600 J·mol^{–1}[/latex]. The infinite-dilution values [latex] \bar{H}_1^{\infty} ext { and } \bar{H}_2^{\infty} [/latex] are found from Eqs. (B) and (C) when [latex]x_1 = 0 in Eq. (B) and x_1 = 1[/latex] in Eq. (C). The results are: [latex] \bar{H}_1^{\infty}=420 \mathrm{~J} \cdot \mathrm{mol}^{-1} ext { and } \bar{H}_2^{\infty}=640 \mathrm{~J} \cdot \mathrm{mol}^{-1} [/latex]

Exercise: Show that the partial properties as given by Eqs. (B) and (C) combine by summability to give Eq. (A) and that they conform to all requirements of the Gibbs/Duhem equation.

Question 10.4: The enthalpy of a binary liquid system of species 1 and 2 at...

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