Liquid-Liquid Solutions

We consider the thermodynamics of two-component systems. The ideas discussed here are easily generalized to multicomponent systems. For a solution consisting of n₁ moles of component 1 and n₂ moles of component 2, the Gibbs energy is a function T and P and the two mole numbers n₁ and n₂. The dependence on these variables is indicated by writing G = G(T,P,n₁,n₂). The total derivative of G is given by

IF the composition of the solution is fixed then we have dn₁ = dn₂ = 0, and the last two terms are zero. In this case the functional form of the Gibb's energy is exactly the same as we have seen previously

where

The chemical potential is defined as

for component 1 and an analogous equation holds for component 2. In general there may a greater number of components and each will have an associated chemical potential that is the derivative of the Gibbs energy with respect to the mole number of that component. It is also evident that the chemical potential is a molar Gibbs energy for one component and for more than one component it is a partial molar Gibbs energy. This is an intensive property and is just the Gibbs energy per mole. For a binary solution the Gibbs energy is

dG = -SdT + VdP + m₁dn₁+m₂dn₂

At constant T and P we have

dG = m₁dn₁+m₂dn₂

A general expression for the Gibbs energy is

G = m₁n₁+m₂n₂

For a one component system G = mn consistent with the statements made previously that m is a molar Gibbs energy.

Other thermodynamic quantities have associated partial molar values. The easiest to see physically is the partial molar volume V_j = (¶V/¶n) _j. Consider the expression analogous to the Gibbs energy above

V = V₁n₁ + V₂n₂

For example, when 1-propanol and water are mixed, the final volume, V of the solution is not equal to the volumes of pure 1-propanol and water. The mixture of two components that can interact in a non-ideal fashion leads to a solution volume that is greater or less than that of the pure components. The partial molar volumes allow this to be quantified.

Other thermodynamic quantities can also be expressed as partial molar derivatives. In general for the jth component we have

The Gibbs-Duhem Equation

Starting with G = m₁n₁ + m₂n₂ we can differentiate to obtain

dG = dm₁n₁ + dm₂n₂ + m₁dn₁+ m₂dn₂

Comparison with the above equation dG = m₁dn₁+ m₂dn₂

leads to

dm₁n₁ + dm₂n₂ = 0.

If we divide both sides by n₁ + n₂ we have

dm₁x₁ + dm₂x₂ = 0

where x₁ and x₂ are mole fractions.

These last two equations are two forms of the Gibbs-Duhem equation. The Gibbs-Duhem equation is important because it tells that if we know the chemical potential of one component as a function of composition, we can determine the other.

For example, the chemical potential of substance 1 in a two component mixture is

Where 0 £ x₁ £ 1. The superscript * is the IUPAC notation for a property of a pure substance. We can differentiate with respect to x₁ and substitute into the Gibbs-Duhem equation to obtain

and since dx₁ = -dx₂ we have

or

Thus, we have shown that one can derive the chemical potential of substance 2 from substance 1.

The expression m_j = m_j* + RT ln x_j implies that we can determine the chemical potential of any subtance based on the knowledge of the chemical potential of the pure substance and the mole fraction x_j. We seek to prove this in the following.

Recall that if two phases are in equilibrium their chemical potentials are equal. We can use this fact to our advantage. At any given temperature a liquid has a vapor pressure. This means that the chemical potential of the vapor above the liquid must equal the chemical potential of the liquid itself. This is just another way of saying that the liquid and its vapor are in chemical equilibrium.

m_j^sln = m_j^vap

If the pressure of the vapor phase is low we can consider it to be ideal. Thus we have

m_j^sln = m_j^vap = m_j^o (T) + RT ln P_j

where we have simply stated the chemical potential of the jth component of the liquid relative to that of its standard state of 1 bar of pressure m_j^o (T). For pure component j the equation becomes

m_j^*(l) = m_j^*(vap) = m_j^o (T) + RT ln P_j^*

Thus

m_j^sln = m_j^* + RT ln P_j/P_j^*.

This is a central result for the study of liquid. This result uses information from the vapor phase chemical potential above the liquid to give us information on the chemical potential in the liquid.

Ideal Solutions

Raoult's law states

P_j = x_jP_j*

where P_j* is the vapor pressure of pure component j. The vapor pressure of component j in an ideal solution is given by the product of its mole fraction and P _j*.

The chemical potential can be expressed as:

m_j = m_j^* + RTln(P_j/P_j^*)

where P_j^* is vapor the pressure of the pure component j in the standard state. The significance of this expression is that we can now consider the equilibrium between vapor and solution to write:

m_j^soln = m_j^vap = m_j⁰ + RTln(P_j/P_j⁰)

but for the vapor P_j⁰ = 1 bar and so

m_j^soln = m_j^vap = m_j⁰ + RTln(P_j)

In the limit that the vapor becomes the pure vapor we have:

m_j^soln = m_j^* + RTln(P_j/P_j^*)

keeping in mind the notation * means the pure component. This is central equation of binary solution mixures.

Using Raoult's law we see that the chemical potential can be expressed as:

m_j^soln = m_j^* + RTln(x_j)

These equation defines an ideal solution.

Two-component phase diagrams

The total vapor pressure over an ideal solution is given by

P_total = P₁ + P₂ = x₁P₁^* + x₂P₂^* = (1 - x₂)P₁^* + x₂P₂^*

= P₁^* + x₂(P₂^* - P₁^*)

A plot of the total pressure has the form of a straight line.

Consider the example in the book of 1-propanol and 2-propanol, which have P₁^* = 20.9 torr and P₂^* = 25.2 torr, respectively. So in this example, the phase diagram has the appearance:

where x₂ is the mole fraction of component 2 (here 2-propanol). The value of the mole fraction in the vapor is not necessarily equal to that of the liquid. In the vapor phase the relative numbers of moles is given by Dalton's law. Applying Dalton's law we find:

y₁ = P₁/P_total = x₁P₁*/P_total.

Or

y₂ = P₂/P_total = x₂P₂*/P_total.

Because of the this the vapor line may not be the same as the liquid line. This is shown in the Figure below.

The purple line was calculated using the Dalton's law expression. What lies between the blue and purple lines? This is the two phase region. If we pick a composition and pressure that is inside this region then we can use a tie line to indicate the composition of each phase.

The tie line shown in Figure (red line) is at a total pressure of 30 torr. You can read the x₂ and y₂ values from the plot (or calculate them using the equations above used to generate the blue and purple curves in the composition-pressure plot.

Temperature Composition Diagrams

We display the composition of the solution at various temperatures in a temperature-composition diagram. To constract a temperature-composition diagram we choose some total ambient pressure such as 760 torr and write

Determine the boiling points of pure 1 and 2 and then begin at some intermediate temperature. Using the Clausius-Clapeyron equation or some empirical relation determine P₁* and P₂* at the intermediate temperature. Then calculate x₁. Calculate y1 using Dalton’s law.

To see an illustration of this procedure read pages 974 – 975 in M&S.