Liquid-Liquid Equilibrium

Background 1: Ternary LLE diagrams

The diagram in its right angled form shows the concentration of two species, in this case mol fraction of species 1 horizontally and of species 3 vertically. Species 2 is not given explicitly but can be calculated from the unity sum of the 3 fractions.

Recall that the three component liquid-liquid eqilibrium system at fixed temperature and pressure (always stated or implied) has, by the Phase Rule, only a single degree of freedom. The classic `triangular diagram' can in fact be regarded as a plot of three separate quantities against this single degree of freedom, which can be taken up by choosing the mol fraction of any one component in either phase.

In the diagram discussed, species 1 and 2 are partially miscible nad species 3 is completely miscible with both.

Consider the single independent variable to be the mol fraction of species 1, plotted along the horizontal axis.

Consider the left hand side of the 2-phase envelope, from its intersection with the horizontal axis to the plait point. It is effectively a plot of the solubility of species 1 in species 2 (which predominates) as the amount of species 3 is varied. The intersection with the horizontal axis, which represents binary mixtures of 1 and 2, is the solubility of species 1 in species 2. We will call this species 2 rich phase the x phase.

Correspondingly, the right hand side of the envelope represents the solubility of species 2 in species 1 as a function of the concentration of the third component. This will be refered to as the y phase.

The third piece of information on this diagram is the relationship between the concentration of species 3 in a species 1 rich y phase (right hand side) to that of species 3 in a 2 rich x phase (left hand side). This is given by the tie lines.

Background 2: Nonideal liquid mixtures

Species which have limited mutual solubility in the liquid phase exhibit positive deviations from Raoult's Law. (Substances which form ideal mixtures are by definition completely miscible.) These are substances whose molecules repel each other.

The quantitative measure of nonideality is the liquid activity coefficient . It is a function of phase composition and temperature. For a component in a mixture which behaves ideally, it will be unity. In vapour liquid equilibrium this modifies the ideal `k-value' and in the present case of liquid-liquid equilibrium it defines the equilibrium between two liquid phases. There will be a value of for each component in each phase, and these will in general all be different.

If we identify the phases as the `x' and `y' phases where the respective mol fraction concentrations are given by these symbols the equilibrium condition is:

^y_i y_i =

^x_i x_i The physical chemistry and thermodynamics of the liquid phase are complicated and will not be discussed in detail here. There are a number of models for the liquid phase which lead to expressions for

of varying complexity. The simplest, which is used here, is the one parameter Margules model. This gives the activity of species 1 of two in a binary mixture in a phase as: RT ln

₁ = A x²₂ A is a characteristic constant for the mixture. Similarly for species 2: RT ln

₂ = A x²₁ Note the subscripts; When the mol fraction of species two x₂ is zero ln

₁ for species one is zero and so

₁ is unity, and vice versa. Thus the pure species one behaves ideally, as would be expected.

The largest absolute value of ln₁ occurs when x₂ is one. This corresponds to a zero concentration of species one, which is said to be at infinite dilution. Thus substances behave most nonideally when they are present in only trace quantities.

The above expressions apply to the `x' phase. Two similar equations can be written for the `y' phase. Given the solubility of species 1 in species 2, which we will call x^o₁ and which on the triangular diagram is at the left hand side of the two phase region on the horizontal axis, we know that the equilibrium pahse will be y^o₁ at the other side of the two phase region. These enable us to determine the constant A for the mixture.

We can write two sets of equilibrium relations, one for each component, and we have two known equilibrium mol fractions in each. Because the equations for the activity coefficients have only a single adjustable parameter, all the equations can only be consistent if the solubility of species 1 in species 2 is the same as that of species 2 in 1, i.e. if y^o₂ = x^o₁.

Note that if the substances are very nearly insoluble then the `y' phase is nearly all species 1, so y^o₁~1 and so is ^y₁. The `x' phase correspondingly is nearly all species 2 and so also x^o₂~1.

A quick estimate of A may be made:

^y₁ y₁ =

^x₁ x₁
1 x 1 =

^x₁ x^o₁

^x₁ = 1/ x^o₁
RTln

^x₁ = A(x^o₂)² ~ A So: ln

^x₁ = - ln x^o₁ = A/RT If the above approximations do not apply then the equation to be solved is more complicated, see below, but it turns out that 1/A is given by the logarithmic mean of either x^o₁ and (1-x^o₁) or of y^o₁ and (1-y^o₁).

Assumptions

Species 1 and 2 are partially miscible and have activities given by the one parameter Margules equation.
Species 3 forms ideal binary mixtures with both 1 and 2.

Calculation Procedure

This outlines the procedure used to calculate points on the two phase envelope and the tie line compositions. More details of the theory and derivations not given above are in the next section.

Given solubility of species 1 in 2, or of 2 in 1, determine y^o₁ (= x^o₂)
Determine; A = (ln y^o₁ - ln (1-y^o₁) ) / (y^o₁ - (1-y^o₁))
Determine plait point compositions: y^p₃ = 1 - (2/A)
y^p₁ = y^p₂ = x^p₁ = x^p₂ = (1-y^p₃)/2
For y₁ from y^p₁ to y^o₁ ....
- Solve:
  A (2y₁ - 1 -y₃) = ln (y₁/(1-y₁-y₃)
  y₃ must lie between 0 and y^p₃.
This determines the right hand boundary of the 2-phase region.
Determine the left hand boundary from the values obtained above since;
x₃ = y₃
x₁ = y₂ = 1 - y₁ -y₃

Further Theory and Derivations

Still under construction.