As in Eq (7.21) for the enthalpy, the molar Gibbs free energy of a solution (g) can be written in terms of pure-component contributions (gA and gB) and an excess value (gex). However, an important contribution needs to be added. For a binary solution, the terms contributing to g are:
gex contains the effects of solution nonideality. The last term on the right hand side arises from the entropy of mixing, and is present in ideal as well as nonideal solutions. According to the definition of g = h -Ts, the Gibbs free energies in Eq (7.33) can be expressed in terms enthalpy and entropy. In particular,
because Ahmjx = 0 (all nonideal behavior is included in gex ).
Asmix for an A-B gas mixture was derived in Sect. 7.2.2. Here Asmx is derived for solid solutions by a totally different method. The starting point is Boltzmann's famous equation inscribed on his tombstone (Fig. 1.6). For a mole of solution, it is
The relation between the gas constant, Boltzmann's constant, and Avogadro's number is R = kNAv. Equation (7.35) represents the difference between the entropy of the solution and the entropy of the pure components. W is the number of different arrangements of the atoms on the lattice sites of the crystal. For the pure constituents, W = 1 because every atom is localized on a site and the atoms are indistinguishable (i.e. they cannot be labeled A1, A2 ).
In order to determine W for the solution, the system consists of N lattice sites on which Na and NB atoms of A and B, respectively, are distributed at random. There are no defects in the crystal, so N = NA + NB.
Starting with an empty lattice, A atoms are added one at a time and the number of ways that each can be placed on the unoccupied sites is counted. The first A atom can be placed on any of the N lattice sites; the second A atom has N - 1 sites available for placement; the third N - 2 available sites, and so on. Since each placement is random, the total number of ways that the entire batch of A atoms can be placed on the available sites is:
However, this method overcounts the total number of ways because it implicitly assumes that the A atoms are distinguishable. For example, the configurations of the first three atoms placed on a six-site crystal are shown in Fig. 7.7. The top line shows two
Was this article helpful?