# Doesn't the eutectic diagram violate Gibbs's phase rule?

EDIT: As the question hasn't attracted any answers, I tried to put in more detail.

Gibbs's phase rule for a 2 component system at constant pressure says that $$F=3-P$$, i.e., that the number of degrees of freedom is three minus the number of phases present. This should mean that when $$P=3$$, there should be no degrees of freedom and three phases should only be in equilibrium at a single point.

However, looking at the eutectic phase diagram, this seems not to be true. Even visually - four lines intersect at the eutectic point, not three.

And my understanding is that the three phases, $$\alpha$$, $$\beta$$ and $$L$$ can be in equilibrium along a line at the eutectic temperature, not only at a single point.

This can be seen by looking at the Gibbs free energy of the three phases at the eutectic temperature, which should look like this:

By the common tangent construction, the three phases are at equilibrium along a range of compositions, and thus $$F=1$$, not $$0$$.

Is there a mistake in my understanding, or is it a 'violation' of the phase rule?

• At constant pressure, F = C - P +1 and there is only one point where all three phases coexist. You can't vary the composition away from the eutectic point and keep all 3 in equilibrium. Jan 31 at 14:20
• To add on to what Jon Custer has mentioned, any system that lies along this "eutectic line" will instead segregate into two phases, one eutectic and the other α/β. This can be understood from the common tangent construction, which shows you that, for a given composition, there are two local minima in free energy corresponding to the two phases. Minimizing free energy involves choosing the proportion of the two phases subject to the constraint of the total composition of your system. Feb 8 at 17:00