What causes the melting point depression known as a eutectic point?

The temperature depression of a eutectic point can be calculated from the enthaplies and entropies of fusion of the two substances.

  • $\begingroup$ Well, the simple answer would be that, because of the relevant enthalpies and entropies the liquid phase has a lower Gibbs free energy than the tie line connecting the solid phases. Was there a deeper question hidden in there? $\endgroup$
    – Jon Custer
    Jul 29, 2015 at 17:57

1 Answer 1


A somewhat longer answer, since I'm afraid my comment may have seemed a bit abrupt...

Lets look at a fairly simple thermodynamic system, the Ag-Ge binary phase diagram. This consists of 3 phases only, fcc Ag, diamond cubic Ge, and the liquid. Taking the published thermodynamic model from J. Wang et al. in Thermochimica Acta 512 240-246 (2011), one can calculate the free energy of each phase as a function of temperature and composition, and then extract the tie lines representing the lowest free energy for the system at each temperature.

Ag-Ge at 900K This figure shows, on the top graph, the free energies in the Ag-Ge system evaluated at 900K. The bottom graph is the calculated phase diagram up to that point. The blue line is the free energy of fcc Ag with Ge as a solute. The steep red line on the right is the free energy of diamond cubic Ge with Ag as a solute. The green line is the free energy of the Ag-Ge liquid phase. The one tie line that is present at this temperature is the nearly horizontal red line (OK, I should pick better colors). This is the common tangent between the fcc Ag(Ge) phase and the diamond Ge(Ag) phase. No other phase has a free energy below this line, so across this range the lowest free energy of the system is a mixture of fcc Ag(Ge) and diamond Ge(Ag).

Ag-Ge at 950K Now we continued the calculation to show, in the upper graph, the calculated free energy curves at 950K. The liquid free energy has now dropped below the tie line between the two solid phases. Thus, the liquid is now a stable phase, and we have two tie lines. One connects the fcc Ag(Ge) phase with the liquid at one composition, and the other connects the diamond Ge(Ag) phase with the liquid at another composition. The liquid has now made an appearance on the phase diagram, with a eutectic occurring at ~921K and ~24 at.% Ge. Right at the eutectic point, the liquid free energy curve touches the tie line between the two solid phases.

A classic text on the thermodynamics of binary systems is Porter and Easterling, Phase Transitions in Metals and Alloys.

  • $\begingroup$ Beautiful diagrams! A tie line is an isotherm? The nearly horizontal red line, which is the common tangent between the fcc Ag(Ge) phase and the diamond Ge(Ag) phase, is not a traditional phase boundary. Rather it is a pseudomorph phase boundary. Is that a common feature of eutectics: the intersection of the free energy lines for a pseudomorphic phase boundary and its corresponding liquid? $\endgroup$
    – Dale
    Jul 29, 2015 at 18:41
  • $\begingroup$ The tie line is not a boundary. It indicates that the free energy along its length is a linear combination of the free energies of the end points. A tie line is a common tangent between two free energy curves. If no other free energy curve goes below the common tangent, than it is the lowest free energy state, and hence the equilibrium state of the system. The points of common tangency are the phase boundaries. $\endgroup$
    – Jon Custer
    Jul 29, 2015 at 18:48

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