becomes:
(3.49)
How do we solve the pressure integral above? One approach is to assume that H2O is an ideal gas.
For an ideal gas:
so that the pressure integral becomes:
Steam is a very nonideal gas, so this approach would not yield a very accurate answer. The concept of fugacity provides us with an alternative solution. For a nonideal substance, fugacity bears the same relationship to volume as the pressure of an ideal gas. Hence, we may substitute fugacity for pressure so that the pressure integral in eqn. 2.130 becomes:
where we take the reference fugacity to be 0.1 MPa. Equation 3.49 thus becomes:
(3.50)
We can then compute fugacity using eqn. 3.44 and the fugacity coefficients in Table 3.1.
Using the data in Table 2.2 and solving the temperature integral in 2.130 as usual (eqn. 2.132), we calculate the ΔGT,P is 3.29 kJ. As it is positive, the left side of the reaction, i.e., brucite, is stable.
The ΔS of this reaction is positive, however, implying that at some temperature, periclase plus water will eventually replace brucite. To calculate the actual temperature of the phase boundary requires a trial and error approach: for a given pressure, we must first guess a temperature, then look up a value of φ in Table 2.1 (interpolating as necessary), and calculate ΔGr. Depending on our answer, we make a revised guess of T and repeat the process until ΔG is 0. Using a spreadsheet, however, this goes fairly quickly. Using this method, we calculate that brucite breaks down at 660°C at 200 MPa, in excellent agreement with experimental observations.
3.6.4 Excess functions
The ideal solution model provides a useful reference for solution behavior. Comparing real solutions with ideal ones leads to the concept of excess functions, for example:
(3.51)
which can be resolved into contributions of excess enthalpy and entropy:
(3.52)
The excess enthalpy is a measure of the heat released during mixing the pure end-members to form the solution, and the excess entropy is a measure of all the energetic effects resulting from a nonrandom distribution of species in solution. We can express excess enthalpy change in the same way as excess free energy:
(3.53)
But since ΔHideal mixing = 0, ΔHexcess = ΔHreal; in other words, the enthalpy change upon mixing is the excess enthalpy change. Similar expressions may, of course, be written for volume and entropy (bearing in mind that unlike volume and enthalpy, ΔSideal is not zero).
Combining eqn. 3.46 with eqn. 3.48 leads to the following:
(3.54)
We can rewrite this as:
(3.55)
Equation 3.55 shows how activity coefficients relate to Henry's and Raoult's laws. Comparing eqn. 3.55 with eqn. 3.39, we see that in the region where Henry's law holds, that is, dilute solutions, the activity coefficient is equal to Henry's law constant. In the region where Raoult's law holds, the activity coefficient is 1 and eqn. 3.55 reduces to eqn. 3.26 since RT ln λi = 0.
Since we know that
comparing equations 3.51 and 3.55, we find that:
(3.56)
which is the same as:
(3.56a)
so that the molar excess free energy associated with component i is simply RT times the log of the activity coefficient. The total molar excess free energy of the solution is then:
(3.57)
We will see the usefulness of the concept of excess free energy shortly when we consider activities in electrolyte solutions. It will also prove important in our treatment of nonideal solid solutions and exsolution phenomena in the next chapter.
Depression of the melting point
In northern climates, salting roads and sidewalks to melt snow and ice is a common practice in winter. We have now acquired the thermodynamic tools to show why salt melts ice, and that this effect does not depend on any special properties of salt or water. Depression of the melting point by addition of a second component to a pure substance is a general phenomenon. Suppose that we have an aqueous solution containing sodium chloride coexisting with pure ice. If the two phases are at equilibrium, then the chemical potential of water in ice must equal that of water in the solution:
(3.58)
(we
