# At the eutectic point, what is the effect on the melting point of adding impurities?

At the eutectic point, what is the effect on the melting point of adding impurities?

1. Increase the melting point
2. Decrease the melting point
3. No effect on melting point

I know adding impurities decreases the melting point of a compound in comparison to the pure compound. But, the eutectic point is the lowest temperature where the new compound (the compound + impurity) melts. I'm indecisive between 1 and 3.

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– user15489
Jun 13, 2015 at 10:38
• Thank you for adding your thinking - the question is much better (and quite interesting).
– user15489
Jun 13, 2015 at 11:01
• Phase diagrams for ternary mixtures are pretty complex (the impurity being the third member). See demonstrations.wolfram.com/BasicTernaryPhaseDiagram, uwgb.edu/dutchs/Petrology/Teutect.htm and tulane.edu/~sanelson/eens212/ternaryphdiag.htm, for example. That said, an impurity that is insoluble in the mix might have no appreciable effect on the MP. Jun 14, 2015 at 4:23

Lets state my assumption that we can represent the free energies of the mixtures in fairly regular thermodynamics of phases ways (as I will get to here quickly). I will also assume fairly limited solid solubilities (think something like the Au-Si system).

OK, just take a system of A and B. The Gibbs free energy of a binary mixture (either solid or liquid) is:

$G(T) = x_{A}G_{A}^{sol/liq} + x_{B}G_{B}^{sol/liq} + x_{A}x_{B}H_{AB}^{sol/liq} + RT(x_{A}\ln x_A + x_{B} \ln x_{B})$, with $x_{A} + x_{B} = 1$, $G_{A,B}^{sol/liq}$ being the Gibbs free energy of pure A (or B) in the solid (or liquid) at that temperature, and $H_{AB}^{sol/liq}$ being the enthalpy of mixing of A and B.

For a eutectic, the free energy of the liquid drops below the tie-line connecting the free energies of the solid phases, and it clearly does this in such a way that the eutectic point has a composition that is intermediate to the compositions defining the solid tie-line. In a perfectly symmetric and regular system, this puts the eutectic at $x_{A} = x_{B} = 0.5$, but at least keep in mind that liquid composition has reasonably large values of $x_{A}$ or $x_{B}$, certainly larger than the minority composition in either solid end-member.

Now, we add a third element. The Gibbs free energy formula becomes longer, since it now takes in to account the binary and ternary interactions. So, one gets:

$G(T) = x_{A}G_{A}^{sol/liq} + x_{B}G_{B}^{sol/liq} + x_{C}G_{C}^{sol/liq} + x_{A}x_{B}H_{AB}^{sol/liq} + x_{A}x_{C}H_{AC}^{sol/liq} + x_{B}x_{C}H_{BC}^{sol/liq} + x_{A}x_{B}x_{C}H_{ABC}^{sol/liq} + RT(x_{A}\ln x_A + x_{B} \ln x_{B} + x_{C} \ln x_{C})$, with $x_{A}+ x_{B} + x_{C} = 1$. This is pretty ugly, but we just added a little bit of C, so $x_{C}$ is small.

Now, back to the assumptions on the system being considered. For the liquid, $x_{A}$ and $x_{B}$ are reasonably large ($\sim 0.5$). In solid A, $x_{A}$ is near 1, $x_{B}$ is small(ish), and $x_{C}$ is really small. In solid B, $x_{B}$ is near 1, $x_A$ is small(ish), and $x_{C}$ is small. Really small.

What has really changed? Well, for $x_{C}$ being really small, you can throw out all terms linear in C. The only thing left (!) is the entropy of mixing terms, where adding a little C results in a change of $RT x_{C} \ln x_{C}$. But, this gets applied to both solids (A and B), and also to the liquid. All the free energy curves shift down by the same amount on an absolute scale. By geometry this means that the tie line connecting the two solid phases and the liquid at the eutectic point does not shift to first order in $x_{C}$.

The eutectic point will not shift.