Here is a path back to the first page of the tutorial and this link leads to the glossary.
At this point we need to step aside a minute and consider the apparently unrelated issues of how many phases can coexist at equilibrium and how many variables need to be fixed to determine the state of the system. We have shown already that in a binary system we can have one-phase regions where the phase can have any composition at fixed P and T or two-phase regions where both phase compositions are fixed. This suggests that actually the number of phases and the number of free variables is related. This relationship is explicitly captured by the Gibbs Phase Rule, which relates the variance (f) of an assemblage to the number of phases , number of components, c, and other restrictions imposed. The phase rule is best understood by thinking of the conditions of equilibrium as a set of simultaneous equations that nature must solve. We learn in elementary algebra that if we have the same number of unknowns (variables) and equations (constraints) then we should expect to find a unique solution, i.e. there are no remaining degrees of freedom in the unknown variables. On the other hand, if we have more unknowns than equations, there is usually a family of solutions, and the dimension of the solution space is the number of extra unknowns. Finally, if there are more equations than unknowns, then unless we are very lucky there will be no solution at all. In the case of the phase rule, the unknowns are P, T, and (c-1)* independent phase compositions. The constraints are the (-1)*c statements of equality of chemical potential among phases. Hence the dimension of the allowed solution space is f = 2 + (c-1)* - (-1)*c = c + 2 - . Once it is given that we have an equilibrium assemblage in a system of c components containing particular phases, the variance is the number of remaining variables we have to fix to determine the state of the system.
Consider, e.g., the one component system H2O. If one phase, say liquid water is present, then f = 1 + 2 - 1 = 2, and indeed we can freely vary both temperature and pressure within a two-dimensional but bounded stability region without change of phase. However, if we insist that water and ice are coexisting (so f = 1 + 2 - 2 = 1) and we fix one more variable say make the pressure 1 atm then the temperature becomes fixed (it must be 0 °C). Ice+liquid in the H2O system is an example of a univariant assemblage, which is restricted to a one-dimensional array (i.e., a line or curve) in P-T-X space. Furthermore, although it is slightly more remote from everyday experience, there is a single point in (P,T) space at P = 6 mbar, T = 0.01 °C where ice, liquid water, and water vapor (steam) can all coexist. This is called the triple point and is an example of an invariant condition f = 1 + 2 - 3 = 0 where merely declaring the number of phases and components has entirely determined the state.
There can be other constraints in the phase rule that lower the variance without adding more phases. For example, if we require that two phases are equal in composition or that three phases are collinear in a three-or-more component system, this takes away one degree of freedom. Hence in the binary system if a solid phase coexists with a liquid of the same composition as the solid (a congruent melting point) this assemblage is univariant: f = 2 + 2 - 2 - one extra restriction = 1. The general term for equilibria with additional restrictions like this is degenerate.
Now let us add a third phase, C, and consider the possibilities. The phase rule tells us that if A, B, and C all coexist in a binary system then the variance is one, so this should only be possible along a line or curve in P-T space. Let's begin at a random (P,T) point off this line and draw the -X diagram at this pressure and temperature. We might find a configuration like that in Figure 4a: the curve for C(X) is everywhere above the envelope of stability defined by the curves for A and B and their common internal tangent segment. Hence C is not stable at any bulk composition at this P and T; its existence has not changed the stability fields at all. Note that we can still find common tangents (shown in yellow) between A and C or between B and C, but these phase pairs are metastable because they always have higher than the A+B mixture.
Now let us imagine that phase C has a higher specific entropy than phases A or B (perhaps C is a liquid, for example). We showed above that the temperature derivative of at constant pressure is -S. So if we raise the temperature and draw a new -X plot, the curve for phase C will have moved down relative to the others (they all move down in absolute terms, but C moves down the most). As we keep raising the temperature at this pressure we will eventually reach a particular, special temperature at which the diagram looks like Figure 4b. Here phases A, B, and C all share a common tangent. It is common for two curves to have a common tangent, but it requires a special geometry to get three curves to share a tangent. This is a univariant condition, and between XA(B) and XB(A), we can have any combination of A+B, A+C, A+B, or A+B+C without any difference in Gibbs free energy.
Now let us keep increasing the temperature at constant pressure and draw a new -X diagram. The curve for phase C keeps moving down relative to the other two, and now there is no longer a common tangent (Figure 4c). Furthermore, the A-B internal tangent segment is now metastable with respect to the A+C and C+B internal tangent segments and part of the C one-phase region. It is no longer possible to have all three phases together, because we have overstepped the univariant reaction. If we wanted to change the temperature and keep on the univariant, we would also have had to change the pressure, by just the right amount to bring the three phases back onto common tangent. In so doing we could map out a curve in P-T space where the univariant assemblage is stable, a univariant curve. The slope dP/dT we would have to follow to stay on the univariant curve is given by the Clausius-Clapeyron Equation and depends on the entropy and volume differences among the phases.
We can also see that crossing a univariant curve (i.e. going from the configuration in Figure 4a
to Figure 4c by way of Figure 4b) causes a change in the sequence of stable phases across the X-axis.
Thus a univariant curve is also a univariant reaction, in this case A+B=C. This means that on
one side of the curve A+B is stable together, on the other side C is stable but A+B is not, and only
on the curve can all three phases coexist.
If we take the sequences of stable phases along the X-axes of the three plots in Figure 4 and transfer them to an isobaric T-X diagram, filling in the spaces between with the sequences we would see if we moved continuously in temperature, we get Figure 5. The boundaries between the one-phase regions (colored black and labelled in red with the name of the stable phase) and the two-phase regions (colored white and labelled in green with the two coexisting phases) are, at the three temperatures corresponding to the three parts of Figure 4, at the same X values as the grey lines we marked in Figure 4. The univariant reaction is shown as a horizontal red line that touches the three one-phase regions and delineates the boundary between the stable two-phase regions.
Now is a good time to back to the applet and click on a univariant
line (shown in red in the P-T projection at upper left). If you hit the line right-on, then in
the -X diagram at upper right you should see three phases with a common
tangent. You might want to zoom in on the univariant line (click-and-drag in the P-T panel
to define a zoom rectangle; shift-click in the P-T panel to unzoom) so you can do this more
accurately. Try clicking in sequence three points at equal pressure (same X coordinate) that
are below, on, and above the univariant. The changes in the -X diagram
should correspond exactly to what we just discussed, except for a fourth phase sitting somewhere
and not participating in the reaction. Now look at the T-X section at lower left...it should remind
you of Figure 5. The blue line on the T-X section shows the temperature at which you have clicked.
Take a minute and explore moving around on and near the univariant reaction in P-T space, watching
the changes in the -X diagram and the way the stable configuration corresponds
to the blue line on the T-X and P-X sections. Note how as you move along the univariant all the
free energy curves move but they stay on a common tangent.
Another way to get a univariant reaction is called a coincidence, and it forms where two coexisting phases have equal composition. In this case a sequence of three -X diagrams at three different temperatures and the section of a T-X diagram that results from extracting the stable sequences across the X axes are all shown in Figure 6. An example of a coincidence is a congruently melting compound, i.e. one that coexists with a liquid of its own composition along its melting curve. Another is a polymorphic transition such as quartz = coesite. It is also common to have degenerate coincidences at the pure endmembers of a binary system.
Try going back to the applet again (not the simplified one but the one with all four phases). If the scale is still zoomed in on the univariant, unzoom by shift-clicking in the P-T panel. Now zoom in on a green portion of the A=B coincidence curve, perhaps near (P,T) = (192,33). Try the exercise again of clicking three points below, on, and above the green coincidence curve at equal pressure. This should remind you of Figure 6. Explore these relations a little bit. If the metastable (grey) B=C coincidence is visible, click on that (perhaps at (P,T) = (204,45) where stable A=B and metastable B=C cross). Why is the B=C coincidence metastable in this region?
There are two kinds of special points to consider in our binary system. The first arises whenever three-phase univariants cross each other (if at least one phase is involved in both reactions), and the second when a three-phase univariant meets a coincidence involving two of its phases. The first is an invariant point, the second is a singular point.
Let's imagine following along a three-phase univariant reaction in our binary system and watch what happens when a fourth phase gets involved. In Figure 7a, the fourth phase is everywhere above the interior tangent segment defining the univariant assemblage, so it is not stable. However, in Figure 7b, still following the univariant reaction to keep the original three phases lined up on their common tangent, we have reached a point where the fourth phase touches the same tangent, so all four are in mutual equilibrium. This is the invariant point. If we move in any direction, one or more of the phases will move off the common tangent, and at the invariant point the composition of all four phases is fixed: we have no degrees of freedom that allow us to keep this four phase assemblage. Proceeding further in the same direction in P-T space, we find in Figure 7c that as the fourth phase dropped below the common tangent to the other three that our original univariant reaction becomes metastable (indicated by the yellow line segment) because the one- and two-phase assemblages involving the fourth phase now have lower than any combination of the three phases involved in the univariant. The three still share a common tangent (since we kept moving along the Clausius-Clapeyron direction), it just is not part of the envelope of minimum curves and tangent segments anymore. This is a general behavior: at any non-degenerate invariant point, each univariant curve switches from stable to metastable. The locations of the three points selected for Figures 7a,b,c are shown as blue circles in P-T space in Figure 7d: we moved along the BCD univariant through the ABCD invariant point and onto the metastable extension of BCD.
Now here's the remarkable thing: starting from our invariant point, it is always possible to move out along four different univariant reactions, each defined by a distinct subset of three of the four phases (or, alternatively, by the one phase among the four not participating). That is, in a binary system, a four-phase invariant point marks the mutual intersection of four univariant curves. For each subset of three phases, we can write a balanced reaction between them, and knowing their entropies and volumes we can calculate the Clausius-Clapeyron slope of the univariant curve involving these three as it takes off from the invariant point. Only one arrangement of the stable and metastable univariant curves around the invariant point is allowed in a binary system; the topological restrictions on arrangments of stable and metastable reactions are expressed by Schreinemaker's Rules, which we will not elaborate here except to say that they are a geometric consequence of the upward concavity of all the (X) curves.
Now let's see how a singular point works. There are many perspectives on this, and the Applet allows you to see them all. If we follow along a stable coincidence curve in P-T space (the animated sequence in Figure 8 follows the B=C coincidence), we will always have two (X) curves that touch at a single point, and one must be "fatter" than the other. As the (X) curve of a third phase (phase D in Figure 8) moves around, the point of tangency where the fatter of the two coincident phases and the third phase are in equilibrium may shift so that it becomes equal to the coincident point. At this location, the singular point, all three phases are in mutual equilibrium. If we keep going in the same direction, the (X) curve of the third phase will keep shifting in the same sense and it is necessary that the common tangent between the third phase (D) and the fatter of the two coincident phases (B) come down below the coincident point and turn it into a metastable coincidence.
Play the Quicktime movie to see the sequence. As you watch the focus slide along the B=C coincidence, note that the contact point of the B and C (X) curves starts out stable, i.e. part of the lower envelope of curves and tangent segments. Note also that the blue line on the T-X and P-X sections goes through the "nose" where the B and C one-phase fields touch and that this point is both at lower T (on the isobaric section) and at lower P (on the isothermal section) than the univariant BCD (red line on the T-X and P-X sections). Finally, note that the composition of B that coexists with D is to the right of the coincident composition where B and C coexist. Now as we move towards the singular point [at (P,T)=(2,369)], the blue and red lines come together and the composition of B that coexists with D shifts over and crosses the coincident point. As we keep going along the now metastable B=C coincidence, C does not appear in the sequence of stable phases at all because the B-D internal tangent segment is lower in than the metastable coincidence. The composition of B that coexists with D keeps moving to the right. And the temperature and pressure differences between the blue line (following the metastable coincidence) and the red line (marking the univariant that we encountered) increase again, with the blue line staying below the red line -- they did not cross, they just touched and separated.