## Liquid-State Physical Chemistry: Fundamentals, Modeling, and Applications (2013)

### 11. Mixing Liquids: Molecular Solutions

So far, we have discussed pure liquids with increasing complexity. In this chapter we start with solutions and, in particular, with molecular solutions. We first deal briefly with some basic aspects, including molar and partial quantities, perfect, and ideal solutions. Thereafter, nonideal behavior is discussed, based on a treatment of the regular solution model. Finally, some possible improvements are indicated.

### 11.1. Basic Aspects

In this section we iterate and extend somewhat on some of the basic aspects that were introduced in __Chapter 2__. The content of a system is defined by the amount of moles *n*_{α} of chemical species α in the system, often denoted as components, which can be varied independently. A mixture is a system with more than one component. A homogeneous part of a mixture, that is, with uniform properties, is addressed as a phase, while a multicomponent phase is labeled a solution. The majority component of a solution is the solvent, while solute refers to the minority component. We restrict ourselves from the outset to binary mixtures and solutions and use, apart from *n*_{α} for the number of moles, *N*_{α} for the number of molecules and *x*_{α} for the mole fraction of component α. The label 1 refers always to the solvent (e.g., *x*_{1} or 1 − *x*), while the label 2 indicates the solute (e.g., *x*_{2} or *x*). As usual, *N*_{A} denotes Avogadro's constant.

**11.1.1 Partial and Molar Quantities**

We refer for an arbitrary extensive quantity *Z* of a mixture to *molar quantities* *Z*_{m} defined by *Z*_{m} = *Z*/Σ_{α}*n*_{α} or for single, pure component α by . For solutions, it is also useful to discuss the situation with respect to extensive quantities *Z* in terms of *partial* (*molar*) *quantities* *Z*_{α} ≡ (∂*Z*/∂*n*_{α})_{P,T,n}_{′}, where *n*′ indicates all components except for the component α. Generally, for *Z* = *Z*(*P*,*T*,*n*_{α}), we have

__(11.1)__

Since the extensive quantity *Z*(*P*,*T*,*n*_{α}) is a homogeneous function of degree 1 in *n*_{α}, we have by Euler's theorem^{1)}

__(11.2)__

So, we have for d*Z* on the one hand __Eq. (11.1)__, and on the other hand differentiating __Eq. (11.2)__ d*Z* = Σ_{α}*Z*_{α}d*n*_{α} + Σ_{α}*n*_{α}d*Z*_{α}. Hence,

(11.3)

This is the (*general*) *Gibbs–Duhem equation*, which puts a constraint on the possible changes in the properties of solutions.

For the volume, the partial volume is given by *V*_{α} = (∂*V*/∂*n*_{α})_{P,T,n}_{′} and the total volume becomes at constant *T* and *P*, using Euler's theorem,

(11.4)

Similarly, for the Gibbs energy *G* we have the partial Gibbs energy *G*_{α} = (∂*G*/∂*n*_{α})_{P,T,n}_{′}, and since this quantity is equal to the chemical potential defined by *μ*_{α} ≡ (∂*G*/∂*n*_{α})_{P,T,n}_{′}, the total Gibbs energy becomes

__(11.5)__

Note that the partial quantity *G*_{α} and chemical potential *μ*_{α} are equivalent, but that for the Helmholtz energy *F* one would have *μ*_{α} = (∂*F*/∂*n*_{α})_{V,T,n}_{′} ≠(∂*F*/∂*n*_{α})_{P,T,n}_{′} = *F*_{α}.

For the Gibbs energy *G* we further obtain at constant *P* and *T*

(11.6)

This is __the__ *Gibbs–Duhem equation* (at constant *P* and *T*) of which another form is

__(11.7)__

where is the chemical potential of the pure component 1. This relation can be used for a consistency check on experimental data. If the molar Gibbs energy *G*_{m} is known, we can obtain *μ*_{1} and *μ*_{2} by differentiation with respect to *x*_{2}, since

__(11.8)__

and solving *μ*_{1} and *μ*_{2} from __Eq. (11.5)__ and __(11.8)__ results in

__(11.9)__

Thus, if the dependence of *G*_{m} as function of *x*_{2} (or *x*_{1} for that matter) is known (__Figure 11.1__), the behavior of *μ*_{1} and *μ*_{2} can be obtained by differentiation. It will be clear that a similar reasoning applies to all partial quantities, for example, *V*_{m}. It will also be evident that *G*_{m} is less easily accessible experimentally than *V*_{m}.

** Figure 11.1** (a) Calculation of the chemical potentials

*μ*

_{1}and

*μ*

_{2}from the Gibbs energy

*G*

_{m}(although illustrated here for

*μ*

_{1}and

*μ*

_{2}using

*G*

_{m}, this calculation can be made for any partial quantity, remembering that

*μ*

_{α}equals

*G*

_{α}); (b) Pressure diagram for a perfect mixture, showing the total pressure

*P*and the partial pressures

*P*

_{1}and

*P*

_{2}.

**11.1.2 Perfect Solutions**

To describe perfect solutions we start with the perfect gas, and thereafter deal with perfect gas mixtures and generalize to perfect (fluid) mixtures. The *perfect gas* can be defined in several ways. Here, we choose a definition based on the chemical potential

(11.10)

where *P* is the pressure, *P*° is a reference pressure (1 bar), and *μ*° is the value for *μ* = *G*_{m} = *G*/*n* if *P* = *P*°. From this definition we obtain for an arbitrary number of moles *n*, *V* = ∂*G*/∂*P* = ∂(*nμ*)/∂*P* = *nRT*/*P*, the well-known equation of state (EoS) for a perfect gas. Hence

(11.11)

For *perfect gas mixtures* we assume that each component contributes independently from the others to the Helmholtz energy, so that we have

(11.12)

This leads to

(11.13)

where *P*_{α} ≡ *x*_{α}*P* is referred to as the *partial pressure*__ ^{2)}__. The total pressure is accordingly given by

*P*= Σ

_{α}

*P*

_{α}or, for a perfect binary mixture, by (

__Figure 11.2__)

(11.14)

known as *Dalton's law of partial pressures*. For the perfect gas mixture the partial volume becomes *V*_{α} = ∂*μ*_{α}/∂*P* = ∂ ln*P*_{α}/∂*P* = *RT*/*P*, equal for all components α.

To calculate Δ_{mix}*G* for a binary gas mixture at and we subtract *n*_{1}*μ*_{1} + *n*_{2}*μ*_{2} at *P*_{1}′ and *P*_{2}′ from *G*_{mix} to obtain

(11.15)

** Figure 11.2** The vapor pressure of the ideally behaving system for: (a) Propylene bromide/ethylene bromide at 85 °C; (b) Ethylene chloride/benzene at 50 °C; (c) Positively-deviating-from-ideality system carbon disulfide/acetone at 35 °C; (d) Negatively-deviating-from-ideality system chloroform/acetone at 35 °C. Many books have used von Zawidski's data [20], usually without reference. It is therefore appropriate to show some of his original figures.

From this expression we obtain the entropy of mixture Δ_{mix}*S* as

(11.16)

In the special case that , that is, when the original pressures of each gas and are equal to the final pressure *P*′, we have and

(11.17)

A *perfect solution* is defined as having the same Gibbs energy of mixing as the perfect gas mixture for the whole composition range, that is, for one mole

__(11.18)__

Equivalently, we have

__(11.19)__

__Equation (11.19)__ forms the basis for all further treatments. In practice only very few systems behave as a perfect mixture, and perfect behavior usually involves a comparable chemical nature of the solvent and solute, although this provides neither a guarantee, nor is required. Generally, strong deviations are observed. An example of both types of near-perfect solutions is shown in __Figure 11.2__, as are examples of a negative and positive from ideal behavior-behaving systems.

Problem 11.1

Recall that a real gas can be described by the virial expansion

Using the inversion of the virial expansion in terms of *P*, show that the expression for *G* becomes

leading for sufficient low pressure to *μ* = *μ*° + *RT* ln (*P*/*P*°).

Problem 11.2

Show that for a perfect gas the energy *U* depends only on the temperature *T*.

Problem 11.3

Show that for a perfect gas mixture, Δ_{mix}*H* and Δ_{mix}*V* are both identically zero.

Problem 11.4

Verify __Eq. (11.9)__.

Problem 11.5

Suppose that the molar volume of a binary solution of components 1 and 2 is given by . Calculate the partial volumes *V*_{1} and *V*_{2}.

Problem 11.6

Suppose that the partial volume of solute 2 in solvent 1 is given by *V*_{2} = *a* + *b*(1 − *x*_{2})^{2}, while the molar volume *V*_{m} (1) of the pure solvent is given by *c*. Calculate the partial molar volume *V*_{1} using the Gibbs–Duhem equation.