PARTIAL MOLAR QUANTITIES BY DIFFERENTIATION OF J AS A FUNCTION OF COMPOSITION

18.1 PARTIAL MOLAR QUANTITIES BY DIFFERENTIATION OF J AS A FUNCTION OF COMPOSITION

The calculation of partial molar quantities, defined [as in Equation (9.12)] as

@n i T ,P,n j

requires the differentiation of J as a function of composition.

Chemical Thermodynamics: Basic Concepts and Methods, Seventh Edition . By Irving M. Klotz and Robert M. Rosenberg Copyright # 2008 John Wiley & Sons, Inc.

CALCULATION OF PARTIAL MOLAR QUANTITIES AND EXCESS MOLAR QUANTITIES

A particularly simple case is shown in Figure 18.1, in which the volume is a linear function of the mole number of glycolamide in a kilogram of water. In this case, the partial molar volume of solute V m2 is constant and is equal to the slope of the line. The partial molar volume V m2 represents the effective volume of the solute in sol- ution, that is, the increase in volume per mole of solute added. From Equation (9.27), written for the volume function,

(18 :2) The linear dependence of V on n 2 at constant n 1 also indicates that V m1 is constant and

V ¼n 1 V m1 þn 2 V m2

is equal to V m1 † , the molar volume of pure solvent. In general, the value of V m2 is not equal to the molar volume of pure solute. For example, glycolamide (Figure 18.1) has an effective volume V m2 in a dilute aqueous solution of 56.2 cm 3 mol 21 ; the pure solid has a molar volume of 54.0 cm 3 mol 21 . For iodine, V

as a solute in liquid perfluoro-n-heptane is 100 cm m2 3 mol 21 , whereas the solid has a molar volume of 51 cm 3 mol 21 , and the supercooled liquid has a molar volume of 59 cm 3 mol 21 . Hydrogen has an effective volume in aqueous solution, at 101.32 kPa (1 atm) and 258C, of 26 cm 3 mol 21 , in contrast to approximately 25,000 cm 3 mol 21 for the pure gas. Furthermore, V m2 for hydrogen (as well as for many other solutes) varies greatly with solvent; it is 50 cm 3 mol 21 in ether and 38 cm 3 mol 21 in acetone. Even more surprising are the effective volumes of some salts in water. For NaCl, the molar volume of the crystal is

27 cm 3 mol 21 compared with 16.4 cm 3 for V m2 . For Na 2 CO 3 , the molar volume of the pure solid is about 42 cm 3 mol 21 compared with 26.7 cm 3 mol 21 for V m2 ; that is, Na 2 CO 3 dissolved in water has a negative effective volume. These observations show very clearly that V m2 reflects not only the volume of the solute molecule, but also the effect that the solvent – solute interaction has on the volume of the solvent.

Figure 18.1. Linear dependence of volume on concentration for dilute solutions of glycol- amide in water. Based on data of Gucker et al. [1].

409 Partial Molar Volume

18.1 PARTIAL MOLAR QUANTITIES BY DIFFERENTIATION OF J AS A FUNCTION

For most solutions, the volume is not a linear function of the composition; the slope of the volume-concentration curve and the value of V m2 are functions of the compo- sition, as illustrated in Figure 18.2. In this case, it can be observed that the volume does not change by equal increments when equal quantities of solute are added successively to a fixed quantity of solvent.

Thus, let us consider what must be the significance of the slope ( @V=@n 2 ) n 1 ,T,P represented by the dashed line in Figure 18.2. According to the principles of calculus, this slope represents the change in volume per mole of added solute n 2 (temperature, pressure, and moles of solvent n 1 being maintained constant) at a fixed point on the curve—in other words, at some specified value of n 2 . As the mass of solvent is 1 kg, n 2 is numerically equal to the molality m 2 . The value of n 2 must be specified because the slope depends on the position on the curve at which it is measured. In practice, this slope, which we represent by V m2 , as in Equation (18.1), refers to either one of the following two experiments.

1. Measure the change in total volume V of the solution when one mole of solute is added to a very large quantity (strictly speaking, an infinite quantity) of the solution at the desired concentration. Because very large quantities of solution are used, the addition of one mole of solute does not change the concentration of the solution appreciably. As in the description of equilibrium in Chapter 10, we may refer to the “infinite copy model.”

2. Measure the change in total volume V of the solution when a small quantity of the solute is added to the solution. Then calculate the change for one mole (that

is, divide DV by Dn 2 ) as if no change in composition occurred when a whole mole of solute was added. Repeat this procedure, but add a smaller quantity Dn 2 0 of solute and compute DV 0 /Dn 0 2 . Repeat again with a still smaller quantity

Figure 18.2. Nonlinear dependence of volume on molality for dilute solutions of sulfuric acid (H 2 SO 4 ) in water (H 2 O). Based on data of I. M. Klotz and C. F. Eckert, J. Am. Chem. Soc. 64, 1878 (1942).

CALCULATION OF PARTIAL MOLAR QUANTITIES AND EXCESS MOLAR QUANTITIES

of added solute. The limiting value

2 !0 Dn 2 @n 2 T ,P,n 1

gives V m2 . These are two equivalent points of view concerning the meaning of V m2 . Clearly,

V m2 does not correspond to the actual change in volume when a whole mole of solute is added to a limited quantity of solution because V m2 is the increase in V upward along the dotted line in Figure 18.2, whereas the actual volume follows the solid line. To calculate V m2 from the data represented in Figure 18.2, numerical or graphi- cal differentiation is necessary, or the data can be fitted to a polynomial equation as in the procedures described in Appendix A.

Most frequently, volume data for solutions are tabulated as density r as a function of composition. The procedure for obtaining V m2 is illustrated by reference to the densities and weight percent concentrations of ethanol – water mixtures (Table 18.1, Columns 1 and 4 at 258C).

To obtain V m2 , that is, ( @V=@n 2 ) n 1 ,T,P , we need values of V for a fixed quantity n 1 of water and for variable quantities n 2 of ethanol. For this purpose we convert the relative weights given in Column 1 to relative numbers of moles, that is, to n 2 /n 1 in Column 2. The numbers in Column 2 also are the moles of ethanol accompanying one mole of water in each of the solutions listed in Column 1.

From this information and the density (Column 4), we can calculate the volume in cubic centimeters that contains one mole of water. The mass of a solution containing n 2 moles of ethanol and one mole of water is

where M is the molar mass. Hence, the volume per mole of water is

mass

cm 3 (mole H 2 O)

Numeric values for these volumes of solution that contain a fixed quantity (one mole) of water are listed in Column 5. The partial molar volumes can be determined by computing the derivative of the quantity in Column 5, the volume of a quantity of solution containing one mole of water and varying moles of ethanol, with respect to the number of moles of ethanol per mole of water, the quantity in Column 2.

Numeric values for the volumes of solution containing a fixed quantity (one mole) of ethanol are listed in Column 7. The partial molar volumes can be determined by computing the derivative of the quantity in Column 7, the volume of a quantity of solution containing one mole of ethanol and varying moles of water, with respect to the number of moles of water per mole of ethanol, the quantity in Column 3.

TABLE 18.1. Densities and Partial Volumes of Ethanol – Water Mixtures a

Dn 2 cm 3 (mol C 2 H 5 OH) Dn 1 20 0.097761

n 1 C 2 H 5 OH n 2 g cm

cm 3 (mol H 2 O)

a Data from N. S. Osborne, E. C. McKelvy, and H. W. Bearce, Bull. Bureau Stand. 9, 327 (1913).

CALCULATION OF PARTIAL MOLAR QUANTITIES AND EXCESS MOLAR QUANTITIES

One way to perform this calculation is to fit the data to a polynomial by the method of least squares (see Section A.1) and to differentiate that function. A quartic fit of the data for one mole of water and varying moles of ethanol yields the function

V ¼18:187 þ 50:997 n =n

2 1 þ 14:592 (n 2 =n 1 ) =n ) 3 þ 5:9914 (n =n ) 2 4 1 2 1

This equation yields an expression for V m2

V m2 ¼50:997 þ 29:18 n 2 =n 1

2 =n 1 ) 2 þ 23:966 (n

2 =n 2 )

A quartic fit of the data for one mole of ethanol and varying moles of water yields the function

V ¼57:930 þ 16:451 n 1 =n 2 þ 0:2312 (n 1 =n 2 ) 2

1 =n 2 ) 3 þ 0:0004360 (n 1 =n 2 ) 4 This equation yields an expression for V m1

V m1 ¼16:451 þ 0:4624 n 1 =n 2

1 =n 2 ) 2 þ 0:001090 (n 1 =n 2 ) 3

The values of V m2 can also be determined graphically from a chord-area plot (see Section A.2) of the ratio of increments listed in Column 6 versus n 2 /n 1 because

Figure 18.3 shows a curve of the derivative calculated from the quartic best fit of V as a function of n 1 /n 2 , which are points from a numeric differentiation, and chords from a graphical differentiation, which show good agreement among the three methods. In the graphical differentiation, the partial derivative ( @V=@n 2 ) n 1 ,T,P can

be determined from a smooth curve drawn through the chords so that the sum of the areas above the chords is equal to the sum of the areas below the chords. Clearly the graphical differentiation that requires careful plotting on a very large graph to match the precision of the data is less convenient than computer fitting of the volume data to a function followed by differentiation.

The partial molar volumes for water in the ethanol solutions can be calculated by analogous procedures. An interesting alternative method is the tangent method [2].

18.1 PARTIAL MOLAR QUANTITIES BY DIFFERENTIATION OF J AS A FUNCTION

Figure 18.3. Graphical differentiation to obtain V m2 in ethanol – water solution compared with derivative of quartic fit of V as function of n 2 /n 1 and numeric differentiation. The solid curve represents the result of differentiation of a fitted quartic polynomial of V as a function of n 2 /n 1 . The squares represent the results of a numeric differentiation. The horizontal lines represent a graphical differentiation.

Partial Molar Enthalpy Absolute values of partial molar enthalpies cannot be determined, just as absolute

values of enthalpies cannot be determined. Thus, it is necessary to choose some state as a reference and to express the partial molar enthalpy relative to that reference state. The most convenient choice for the reference state usually is the infinitely dilute solution. Without committing ourselves to this choice exclusively, we will nevertheless use it in most of our problems.

The relative values of partial molar enthalpies are used so frequently that it has become customary to use the special symbol L mi to represent them. Thus, L mi , which is the relative partial molar enthalpy, is defined by the equation

mi 8 (18 :7) in which H8 mi is the partial molar enthalpy of component i in the standard state, that is,

L mi ¼H mi

the infinitely dilute solution. For the solvent, H8 m1 equals H † m1 , which is the molar enthalpy of the pure solvent. For the solute, H8 m2 generally does not equal H m2 † , which is the molar enthalpy of the pure solute.

CALCULATION OF PARTIAL MOLAR QUANTITIES AND EXCESS MOLAR QUANTITIES

Relative partial molar enthalpies also can be visualized in terms of a diagram, such as Figure 18.4. Although the absolute position of H ml or H8 ml on the enthalpy scale cannot be specified, the difference between them can be determined.

Enthalpies of Mixing Experimental data from which relative partial molar enthalpies can be calculated

consist of enthalpy changes for mixing processes, which are commonly those that give integral heats of solution. An example of an integral heat of solution is the enthalpy change for the process of dissolving one mole of NaCl in 1000 g (55.51

moles) of pure H 2 O to give a 1-molal solution, as shown by Equation (18.8):

NaCl(s) þ 55:51 H 2 O(l) ¼ solution of 1 NaCl and 55:51 H 2 O(l) (18 :8) For such a process

(18 :9) For any extensive thermodynamic property of a solution [Equation (9.27)]

DH ¼ H final

initial

J ¼n 1 J m1 þn 2 J m2

Thus, DH ¼ H final

2 H m2 † (18 :10) in which H m1 † and H m2 † are the molar enthalpies of pure solvent and pure solute,

initial ¼n 1 H m1 þn 2 H m2

1 H † m1

respectively. As H 8 m1 ¼H † m1 ,

Figure 18.4. Relative partial molar enthalpy.

415 If we add (H8 m2 2 H8 m2 ) to the preceding equation, we obtain

18.1 PARTIAL MOLAR QUANTITIES BY DIFFERENTIATION OF J AS A FUNCTION

(18 :12) Differentiation of Equation (18.12) term by term with respect to n 1 at constant

n 2 yields @DH

@L m1

@L m2 ¼L m1 þn 1 þn 2 (18 :13) @n 1 n 2 @n 1 n 2 @n 1 n 2

From the Gibbs – Duhem equation [Equation (9.33)]

(18 :14) Consequently, the last two terms of Equation (18.13) are equal to zero, so that

If DH is plotted against n 1 at constant n 2 , a graphical differentiation by the chord-area method will yield L m1 as a function of composition. Alternatively, the data could be fitted to a polynomial and the derivative of that polynomial then could be computed.

Differentiation of Equation (18.12) with respect to n 2 at constant n 1 yields @DH

(18 :16) because the Gibbs – Duhem equation eliminates the first two terms of the equation. To

¼L m2

† m2

obtain L m2 from the experimental data, it is necessary to have a value of L m2 † , which is the relative partial molar enthalpy of the pure solute.

we can observe that

From the limiting value of the slope of a DH versus n 2 plot, it is possible to calculate L † m2 . It follows that L m2 at any concentration can be evaluated from

@DH

@DH

L m2 ¼

@n 2 n 1 n 2 !0 @n 2 n 1

CALCULATION OF PARTIAL MOLAR QUANTITIES AND EXCESS MOLAR QUANTITIES

Figure 18.5. Enthalpy of solution of n 2 moles of gaseous HCl in 1 kg of H 2 O. Based on data from Ref. 3.

Lewis and Randall [3] have recalculated some of Thomsen’s data on the heat absorbed when n 2 moles of gaseous HCl is dissolved in 1000 g of H 2 O. A plot of DH as a function of n 2 at constant n 1 is shown in Figure 18.5. To find L m2 it is necessary to calculate ( @DH=@n 2 ) n 1 ,T,P at various molalities of HCl. The data in Figure 18.5 have been fitted to a quadratic polynomial by the

method of least squares. 1 The equation obtained is

2 þ (216:94 + 0:41) (n ) 2 2 Therefore

dDH

lim

¼ 17375 cal mol

n !0

2 dn 2

Hence, the relative partial molar enthalpies of HCl in aqueous solutions can be expressed by the equation

2 þ 17,375 ¼ 432 n 2 ¼ 432 m 2 (18 :19) Thus, for a 10-molal solution, the value of L m2 can be calculated from Equation

(18.19) as

L m2 ¼ [432 cal (mol kg ) ][10 :0 mol kg ] ¼ 4320 cal

1 We can tabulate values of D(DH) /(Dn 2 ), D(DDH ) /(Dn 2 ), and D(DDDH ) /Dn 2 and find that the third quan- tity varies randomly about zero, which is a behavior that indicates that the second derivative of DH with

respect to n 2 is constant within experimental error, or we can fit the data to polynomials of successively higher powers until no significant improvement in standard deviation occurs for the coefficients.

417 Although Thomsen’s data provide a basis for illustrating this method of calcu-

18.1 PARTIAL MOLAR QUANTITIES BY DIFFERENTIATION OF J AS A FUNCTION

lation, they imply a relationship between DH and m 2 at low molalities that is probably not an accurate description of the system. A better functional relationship, which is based on more modern data, will be discussed in Chapter 19.

Enthalpies of Dilution Relative partial molar enthalpies can also be obtained from measurements of enthal-

pies of dilution. Humphrey et al. [4] have used enthalpy of dilution measurements to calculate relative partial molar enthalpies in aqueous solutions of amino acids. Their data for DH dil of aqueous solutions of serine are shown in Table 18.2, where m i is the

initial molality of the solution, m f is the molality after addition of a small amount of solvent, and DH dil is equal to the measured DH divided by n 2 , which is the number of moles of solute in the solutions. If the initial solution contains n 1 moles of solvent and n 2 moles of solute, and if n 1 0 moles of solvent are added,

If we add (H8 m1 2 H8 m1 ) to the first parenthesis and (H8 m2 2 H8 m2 ) to the second parenthesis, the result is

TABLE 18.2. Enthalpies of Dilution of Aqueous Solutions of Serine [4] m i

m f DH dil 3.8317

m f DH dil

CALCULATION OF PARTIAL MOLAR QUANTITIES AND EXCESS MOLAR QUANTITIES

If we assume that L m2 /n 2 can be expressed as a polynomial in the molality m, L m2

¼A m þA m 2 þA m 3 (18 :23)

then DH dil ¼A 1 (m f i ) þA 2 (m 2 f 2 i ) þA 3 (m 3 f 3 i )

(18 :24) The constants in Equations (18.23) and (18.24) can be determined by fitting DH dil to a

function of m f 2m i with a least-squares fitting procedure (Section A.2). The appropriate degree of the polynomial is found by increasing the number of terms as long as the sum of square deviations decreases significantly. The results for the

data in Table 18.2 for a cubic polynomial are as follows: A 1 ¼ 2723 + 5.5 J mol 21 (mol kg 21 ) 21 , A 2 21 (mol kg 21 ) ¼ 137 + 3.4 J mol 22 , and A 3 ¼ 212 J mol 21 (mol kg 21 ) 23 . Then, from Equation (18.23)

L m2 ¼n 2 A 1 m þn 2 A 2 m 2 þn 2 A 3 m 3 (18 :25) But,

(18 :26) where M 1 is the molar mass (in kg) of the solvent. Therefore, for 1 kg of solvent, in

n 2 ¼n 1 M 1 m

which case n 1 ¼ 1/M 1 , L ¼n 1 M 1 A 1 m 2 þn 1 M 1 A 2 m 3 þn 1 M 1 A 3 m 4

¼A 1 m 2 þA 2 m 3 þA 3 m 4 (18 :27) Then,

n 2 T ,P,n 1 @m T ,P,n 1 ¼1=M 1

Of course, the expression for L m2 is independent of the amount of solvent, because it is an intensive variable. Then, for 1 kg of solvent, with n 2 ¼ m and n 1 ¼ 1/M 1 ,

(18 :29) The values of L m1 and L m2 calculated from Equations (18.28) and (18.29) are plotted in

1 (A 1 m 2 þ 2A 2 m 3 þ 3A 3 m 4 )

Figures (18.6) and (18.7).

18.1 PARTIAL MOLAR QUANTITIES BY DIFFERENTIATION OF J AS A FUNCTION

Figure 18.6. Relative partial molar enthalpy of water in serine solutions. Data from Ref. 4.

Some authors (4,5,6,9,10) have used the apparent molar quantity FJ 2 , where

n 2 The apparent molar enthalpy FH 2 has been used to derive relative partial molar

enthalpies from enthalpy of dilution data. The apparent molar enthalpy is defined as

Figure 18.7. Relative partial molar enthalpy of serine in aqueous solutions. Data from Ref. 3.

CALCULATION OF PARTIAL MOLAR QUANTITIES AND EXCESS MOLAR QUANTITIES

and the apparent relative molar enthalpy is

because L m1 † is equal to zero. The procedure that we have described for the treatment of enthalpy of dilution data is more direct, however. In addition, whenever the value

of FJ 2 can be expressed as a polynomial in m, FJ 2 ¼ a þ bm þ cm 2 (18 :33)

the value of J for a solution containing 1 kg of solvent can also be expressed as a polynomial in m, because then n 2 ¼n 1 M 1 m , and n 1 ¼ 1/M 1 .