[Mike Dacre]

I understand why an impure solid will have a decreased melting point, however, in my recent organic chemistry lab section I mixed two finely ground pure solids together and observed that the mixture melted at a lower temperature than either pure solid.

I understand this in principle, but whenever mixed melting point theory is taught, it is taught from the perspective that impurities reduce the strength of a crystal structure. However, in my experiment I had two *solid* pure substances, mixed together. The crystal structure could not possibly be disrupted, as the compounds had never mixed in liquid form. So my question is: how does that work? Is it just the case that a small amount of one or both of the pure substances is melting at the edges of the crystals, and that when this happens, this tiny amount of impure liquid can dissolve more of the two pure solids, and that this is what causes the melting point depression?

[Arkcon]

The crystal structure could not possibly be disrupted, as the compounds had never mixed in liquid form. So my question is: how does that work?

Really?

Is it just the case that a small amount of one or both of the pure substances is melting at the edges of the crystals, and that when this happens, this tiny amount of impure liquid can dissolve more of the two pure solids, and that this is what causes the melting point depression?

Ah, I see you figured it out.

Try not to see it as simply a disruption of crystal structure.  Try also to visualize it as driven by entropy.  Can you make some comparisons between a pure solid, a melted pure solid, and a mixture, in terms of entropy,or disorder?

[Mike Dacre]

The crystal structure could not possibly be disrupted, as the compounds had never mixed in liquid form. So my question is: how does that work?

Really?

I take it from you question that my assumption is wrong 

Is it just the case that a small amount of one or both of the pure substances is melting at the edges of the crystals, and that when this happens, this tiny amount of impure liquid can dissolve more of the two pure solids, and that this is what causes the melting point depression?

Ah, I see you figured it out.

Try not to see it as simply a disruption of crystal structure.  Try also to visualize it as driven by entropy.  Can you make some comparisons between a pure solid, a melted pure solid, and a mixture, in terms of entropy,or disorder?

In order of increasing entropy:

Solid < Pure Liquid < Mixture (generally; it is possible that the intermolecular forces in the mixture could be more 'organizing' than the intermolecular forces of the liquid; but if we assume an ideal mixture, then this progression is true).

So my model makes sense then? In a simple—almost-ideal—mixture, the small amount of liquid forming at the edges of the finely ground crystals are dissolving the two pure crystalline solids, forming a rapidly spreading entropically favored liquid?

Is there anything else going on with the vapor pressure of the solids or anything else?

Also, would it be fair to assume that if we have two pure compounds that are miscible, A and B, and we melt both and then freeze the mixture, that mixed solid will have a lower melting point than the melting point of the two mixed pure solids, which will in turn be lower than either of the single pure solids?

i.e.: Melting point of mixed solid < melting point of two pure solids mixed together < melting point of either solid

Or would the melting point of the mixed solid and the melting point of the two pure solids mixed together be the same? Assuming that both are ground equally.

[Enthalpy]

Mixing powders of two pure solids isn't very efficient to lower the melting point but it helps a bit in metallurgy. Though, as far as possible, the normal procedure would be to melt the most difficult metal first, then mix the easiest in.

You suggested a tiny part melts: this isn't even necessary. Solids get some liquid behaviour well below their melting point:

• They can diffuse in an other. Gold and lead are long known for it even at low temperature. Gold or sodium in silicon too, because silicon is so sensitive to both. That is observable, depending on how sensitive the measure is, hundreds of kelvins below the melting point. It explains the mixe's lower melting point.
• They creep, slowly but more and more as temperature rises, despite the stress is well below the yield stress at that temperature and it's far from melting. This happens typically at gas turbines, where nickel alloys operate at +700°C (some 700K below the melting point), as limited by the deformation speed over months. Providers give creep curves for such alloys, which limit the stress to 1/2 to 1/4 the yield strength.

Such processes, used for instance at sintering, rely on a few lucky atoms that receive many times kT, hence depend very steeply on the temperature.

[Enthalpy]

The lower melting point of allied solids can be understood at molecular scale too.

Take a solid whose constituents need to be well-ordered, like Fe₃C for instance. The liquid accepts each constituent anywhere at every time, so melting keeps equally easy, but the solid accepts only one constituent at each location while each constituent gets less abundent in the liquid mix, so solidification becomes more difficult. This favours the liquid phase, that is, it lowers the melting point.

[mjc123]

Also, would it be fair to assume that if we have two pure compounds that are miscible, A and B, and we melt both and then freeze the mixture, that mixed solid will have a lower melting point than the melting point of the two mixed pure solids, which will in turn be lower than either of the single pure solids?

Depends what happens on freezing, i.e. on what you mean by "miscible". If A and B are miscible in the liquid phase but not in the solid phase, you will get a mixture of solid A and B - an intimate mixture of very small crystals, perhaps, but still, chemically and thermodynamically, two separate solid phases. It will have the same melting point as your physically mixed two solids. (Strictly it doesn't have a melting point, it melts over a temperature range. Have you studied phase diagrams, eutectic mixtures etc. yet?)
If A and B are miscible in the solid phase too, you will get a solid with A and B mixed at a molecular level - a solid solution. This will have a melting point between those of A and B, i.e. higher than the physical mixture. Think about the entropy - a molecularly mixed solid will have higher entropy than two pure solids, so the entropy change on melting (to the same liquid) will be lower. By contrast, the increase in entropy going from large lumps of A and B to intimately mixed very small crystals is absolutely tiny compared with going to molecularly mixed A and B, hence the first answer.

[Mike Dacre]

Also, would it be fair to assume that if we have two pure compounds that are miscible, A and B, and we melt both and then freeze the mixture, that mixed solid will have a lower melting point than the melting point of the two mixed pure solids, which will in turn be lower than either of the single pure solids?

Depends what happens on freezing, i.e. on what you mean by "miscible". If A and B are miscible in the liquid phase but not in the solid phase, you will get a mixture of solid A and B - an intimate mixture of very small crystals, perhaps, but still, chemically and thermodynamically, two separate solid phases. It will have the same melting point as your physically mixed two solids. (Strictly it doesn't have a melting point, it melts over a temperature range. Have you studied phase diagrams, eutectic mixtures etc. yet?)

Yes I have, I understand that part. Thank you.

If A and B are miscible in the solid phase too, you will get a solid with A and B mixed at a molecular level - a solid solution. This will have a melting point between those of A and B, i.e. higher than the physical mixture. Think about the entropy - a molecularly mixed solid will have higher entropy than two pure solids, so the entropy change on melting (to the same liquid) will be lower. By contrast, the increase in entropy going from large lumps of A and B to intimately mixed very small crystals is absolutely tiny compared with going to molecularly mixed A and B, hence the first answer.

This is great, it never occurred to me to consider differences in miscibility in the solid phase vs. liquid phase, I assumed that the two things would always correlate. That makes a lot more sense now, thank you for your explanation.