Department of Geosciences
Department of Geosciences 		Uni Basel
Lecture Notes and Tutorials: Reciprocal Solutions

by Christian de Capitani Navigation Window: Thermodynamics in Basel

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Step 1: Introduction

For the present exercice you should have some experience with the THERIAK-DOMINO software and its database. (A complete description of the database is included in the THERIAK homepage).
The name "reciprocal solution" was originally used for molten salts. It is however equally applicable to any multisite solid solution.
For simplicity we will only discuss models with two sites and two ions per site. Systems with more sites or exchangable ions can be treated exactly in the same way (only more so).

Consider a solid solution with the general formula:
(A,B)(X,Y)

If such a system is considered a mixture of A, B, X and Y, then it is constrained by the requirement, that the concentration of the ions on the first site is equal to that on the second site. Thus

x(A) + x(B) = x(X) + x(Y) = 0.5

In order to avoid this additional constraint, reciprocal solutions are usually described in two different ways:

(1) Mole fractions for each site separately, both having the sum of 1.
(=site occupancies) 		x(A) = site occupancy of A (on site 1)
x(B) = site occupancy of B (on site 1) = 1-x(A)
x(X) = site occupancy of X (on site 2)
x(Y) = site occupancy of Y (on site 2) = 1-x(X)
 
(2) A mixture of the four species AX, AY, BX and BY. x(AX) = concentration of species AX
x(AY) = concentration of species AY
x(BX) = concentration of species BX
x(BY) = concentration of species BY

While in the first case each composition corresponds to a unique set of site occupancies, in the second formulation almost all compositions may be expressed by an infinity of different combinations of the four species.

The difficulty in dealing with reciprocal solutions is that we will use both formulations simultanously and have to be careful in order to stay consistent.

If we know the distribution of species, we can immediately calculate the site occupancies:

x(A) = x(AX) + x(AY)
x(B) = x(BX) + x(BY) = 1-x(A)
x(X) = x(AX) + x(BX)
x(Y) = x(AY) + x(BY) = 1-x(X)

To calculate the concentrations of the species from site occupancies is impossible without further assumptions.

We will first consider an ideal solution consisting of the four species AX, AY, BX and BY as endmembers. We know that at equilibrium all reactions among the species have

and because of ideality the activities a(AX) etc. are equal to the concentrations x(AX) etc.
The only reaction among the four species is the cross-reaction:

AX + BY = AY + BX

thus:


where:
= Gibbs Free Energy of the endmembers
= the chemical potential of the species s in the ideal solution.

summarizing:

The left hand side is G of the endmember reaction where



from this follows:
x(AY)·x(BX) = k1 · x(AX)·x(BY)

Now lets begin with the excercices (Step 2).
This exercices include of the following steps: (You have to complete step 2 in order to continue)

The following steps can only be reached after solving the problem in step 2

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