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I am tutoring my nephew for his science GCSEs; and his chemistry textbook states that you can tell how pure any compound is by how close its melting and boiling points are to the "official" mp and bp for that compound. It says that impurities will always lower the melting point of any compound, but - paradoxically (intuitively speaking) - will always raise the boiling point. But it just states this as a fact and offers no explanation!

I found a simple explanation for why the melting point is reduced in the case of crystalline solids:

Foreign substances in a crystalline solid disrupt the repeating pattern of forces that holds the solid together. Therefore, a smaller amount of energy is required to melt the part of the solid surrounding the impurity. This explains the melting point depression (lowering) observed from impure solids.

... although this doesn't explain why impurities in solids that are not crystalline also lower the melting point.

But I can't find any easy to understand explanation anywhere for why impurities would always increase the boiling point of any compound. I've found explanations that are specific to water (and even then, the explanations I've seen only make sense if the impurity binds more strongly to water molecules than water molecules bond to each other, which isn't always the case); and I found one explanation that says it's because of increased solution phase enthalpy, without explaining what on earth that means!

So I emailed a Professor of Physical Chemistry about this and she replied

The reason for impurities lowering the melting point yet increasing the boiling point is because the impurities stabilise the liquid phase, making it more energetically favourable. This extends the liquid range to lower temperatures (lowering the melting point) and to higher temperatures (raising the boiling point). The liquid phase is stabilised due to the entropy increase when you have the impurity (or any solute) in the liquid. Normally solids do not dissolve the impurities, so the melting point involves transition to pure solid (with impurity separating out or staying in the liquid state).

She made it clear she didn't have time to answer any follow-up questions, and I can't make full sense of her answer, so I am hoping someone here can understand her explanation and elucidate it.

I think I understand why impurities increase the entropy of a substance in its liquid phase by more than they increase the entropy of the same substance in its solid or gasseous phases - I presume impurities don't significantly affect the entropy of a solid because the solid can't rearrange itself into different configurations without absorbing a lot of energy; and they don't affect the entropy of a gas because the gas can disperse over an almost infinite distance to all intents and purposes, so it will be almost infinitely disordered even when pure; but they do increase the entropy of a liquid because there are many more ways of arranging the molecules within a liquid if there is more than type of molecule in that liquid, and the molecules in a liquid can rearrange themselves in all those different configurations without any input of energy.

First of all, have I got that much right?

But even if I have, I don't understand why increasing the entropy of the liquid by adding an impurity to it would make it energetically more stable - and therefore both more difficult to freeze and more difficult to vapourise.

Can anyone help to explain this?

Dave

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All generalities are false (as is this one). Though impurities usually lower the melting point (m.p.) by disrupting crystallization on the atomic order, consider the phase diagram of the binary alloy (amalgam) HgxAg1- x: Pure Hg melts ~-39°C, and adding even a little bit of impurity raises the melting point considerably!

And without resorting to entropic explanations, consider that a high boiling point (b.p.) solute would "hold onto" the solvent, decreasing vapor pressure. $\ce{CaCl2}$ will absorb $\ce{H2O}$ from the air, driving down the vapor pressure (and humidity) even after the $\ce{CaCl2}$ liquefies.

That said, the answer they expect on the exam is that impurities decrease the m.p. and increase b.p.

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  • $\begingroup$ Hi DrMoishe Pippik I can see why that would be the case if the solute binds more strongly to the solvent molecules than the solvent molecules bind to each other, but I don't understand why that would be the case if the impurity binds more weakly to the solvent molecules then the solvent molecules bind to each other? Also, are you saying that impurities in non-crystaline solids do not lower their melting points? Many thanks for your help Dave $\endgroup$
    – Dave Rado
    May 3 at 2:10
  • $\begingroup$ @DaveRado, do you mean why it lowers the m.p.? The impurity gets in the way of crystal formation. $\endgroup$ May 3 at 2:12
  • $\begingroup$ Hi DrMoishe Pippik, No. I am asking why a solute would "hold onto" the solvent, decreasing vapor pressure, if the solute bonds more weakly to the solvent than the solvent molecules bond to each other. I also asked whether you are saying that impurities in NON-crystaline solids do not lower their melting points? Thanks for your help. Dave PS - I tried to use "@DrMoishe Pippik" above but it deleted it every time I saved, so I had to remove the "@" symbol and use "Hi" instead of "@" - any idea why? $\endgroup$
    – Dave Rado
    May 3 at 2:25
  • $\begingroup$ Non-crystalline solids, e.g., glass, do not have clearly defined melting points. Consider that window glass is a mixture of substances dissolved together, so it's melting point is so broad as to be called the "softening point". $\endgroup$ May 3 at 2:27
  • $\begingroup$ Hi DrMoishe Pippik Thanks for the clarification regarding the melting points of non-crystalline solids. However, my main question was about BOILING points. I can understand why a solute would "hold onto" the solvent, decreasing the vapor pressure, if the solute bonds more strongly to the solvent molecules than the solvent molecules bond to each other - but I don't understand why that would still be the case for solutes that bind more WEAKLY to the solvent molecules than the solvent molecules bind to each other. Please could you help me to understand that as well? Many thanks Dave $\endgroup$
    – Dave Rado
    May 3 at 11:37
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I just want to put forward my explanation on this - melting point is conversion of solid to liquid . so, if the impurities are there - it definitely weakens the force of attraction between the molecules which holds them in a regular lattice usually. Hence you need to supply less energy to make it to liquid unlike with the pure substance with regular lattice arrangement. Boiling point is like a surface phenomenon where already the liquid has mobility enough to move randomly and if impurities are also there, it definitely provides hindrance for the molecules to evaporate from the liquid surface. Imagine its just like a marathon..so crowded ! Hence more energy required to overcome the hindrance caused by the impurities. Hence, impurity decreases the melting point and increase the boiling point I guess! Freezing point is to bring the molecules closer. so, if impurities are there - molecules can be made into tightly packed as the impurities will also be present in the interstitial space of lattice that can help to bring it back to tightly packed solid state quickly with less energy.

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    $\begingroup$ Boiling is not a surface phenomena, it is a bulk thermodynamic phase transition. The rest of the answer similarly ignores thermodynamics of mixtures. $\endgroup$
    – Jon Custer
    Nov 10 at 13:54
  • $\begingroup$ Thanks for letting me know the mistake.. But I have a doubt, when we boil, it becomes gas ..so evapoartion is surface phenomenon right? Then why can't the impurities give hindrance to the evaporation - which leads to increase in boiling point? $\endgroup$
    – Velammal
    Nov 12 at 13:01
  • $\begingroup$ When boiling water on the stove, take note of the bubbles coming up from the bottom of the pan. And do not confuse kinetic effects (such as nucleation and surface area) with thermodynamics. $\endgroup$
    – Jon Custer
    Nov 12 at 22:09

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