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Author Topic: Explanation of melting point depression in impure solids  (Read 18384 times)

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Orthocelsus

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Explanation of melting point depression in impure solids
« on: September 05, 2011, 04:36:03 PM »

I wonder if someone could provide an explanation of why the physical addition of an impurity to a pure substance  causes a depression in the mixture's melting point lower than the mp of either the pure substance or the impurity.

I'm not looking for equations, but for a description in terms of atomic/molecular and crystal theory that explains how an impurity extrinsic to the crystal structure of the pure substance can lessen the intracrystalline forces such that the mp is lowered. I can understand how mp depression could occur when an impurity is incorporated into the crystal, by disrupting the intracrystalline forces, but I have a hard time imagining how merely mixing the substance and impurity together would cause this.

Since the resulting impure mixture melts at a temperature below the mp or the impurity, it's not a case of the impurity melting first and dissolving the substance. I've taught a few organic lab course where I taught the important principle of melting point as a criterion of purity, but when asked how this happened, I always came up empty.

And while we're at it, how does adding one type of cooking oil to another raise the smoking point?
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rajajidwivedi

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Re: Explanation of melting point depression in impure solids
« Reply #1 on: September 12, 2011, 05:42:02 AM »

for this first we should understand the m.p. it is the temp at which the vapor pressure of liquid and solid state of compound is equal, means if we draw a curve of vapor pressure of solid and liquid state then point of intersection will be represent the mp. we know that on adding any nonvolatile substance vapor pressure decreases and thus mp decreases.
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fledarmus

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Re: Explanation of melting point depression in impure solids
« Reply #2 on: September 12, 2011, 06:56:40 AM »

for this first we should understand the m.p. it is the temp at which the vapor pressure of liquid and solid state of compound is equal, means if we draw a curve of vapor pressure of solid and liquid state then point of intersection will be represent the mp. we know that on adding any nonvolatile substance vapor pressure decreases and thus mp decreases.

How does the melting point have anything to do with vapor pressures? It is an equilibrium between solid and liquid, not gas

The best physical explanation I've heard is that it has to do with the entropy of the system. If there are defects in the crystal structure caused by the presence of impurities, it takes less energy to break the crystal structure down, and it will melt at a lower temperature.
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Borek

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Re: Explanation of melting point depression in impure solids
« Reply #3 on: September 12, 2011, 08:44:08 AM »

How does the melting point have anything to do with vapor pressures? It is an equilibrium between solid and liquid, not gas

Yes, and approaching it from the vapor pressure side is a little bit awkward - but it gives correct results. rajajidwivedi  is right about it.
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Yggdrasil

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Re: Explanation of melting point depression in impure solids
« Reply #4 on: September 13, 2011, 04:26:23 AM »

I'm not looking for equations, but for a description in terms of atomic/molecular and crystal theory that explains how an impurity extrinsic to the crystal structure of the pure substance can lessen the intracrystalline forces such that the mp is lowered.

You are correct that, if the impurities are not incorporated into the crystal, the intermolecular forces in the crystal do not change.  However, these intermolecular forces are not the only factor that affects the melting point of a substance.  Recall that the melting point of a solid is defined as the temperature at which the solid and liquid phases are at equilibrium. Mathematically, this means that the change in free energy of going from solid to liquid is zero (ΔG = 0).  Now, also recall that ΔG is composed of two components: enthalpy (ΔH) and entropy (ΔS).  Therefore, the melting point not only depends on the strength of the intermolecular forces (taken into account by ΔH), but also the change in entropy associated with the phase transition. 

The relative effects of enthalpy and entropy are described by the equation ΔG = ΔH - TΔS. Since ΔG = 0 at the melting point, we can solve for the melting point: T = ΔH/ΔS.

What happens to this value when we add an impurity into the liquid phase? The change in enthalpy (ΔH) describes the amount of heat required to break the interactions between solid molecules in order to become a liquid. This value does not change much when you add an impurity. The change in entropy (ΔS) describes the entropy gained when going from the much more ordered solid phase to the much less ordered liquid phase. Adding an impurity to the liquid phase increases the entropy of the liquid without affecting the entropy of the solid much; the end result is that ΔS larger for a solid melting into an impure liquid than a pure liquid.

Adding an impurity to the liquid phase causes ΔS to increase without changing ΔH. As you can see from the equation above, this situation must cause the melting point (T) to decrease. Therefore, adding an impurity, will cause a substance to melt at a lower temperature.

In essence, melting is a process that is driven by the increase in entropy associated with going from the solid phase to the liquid phase. By making the liquid phase more disordered, this gain in entropy becomes larger, and melting becomes slightly more favorable.
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Orthocelsus

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Re: Explanation of melting point depression in impure solids
« Reply #5 on: September 26, 2011, 10:05:00 PM »

Quote
Adding an impurity to the liquid phase causes ΔS to increase without changing ΔH. As you can see from the equation above, this situation must cause the melting point (T) to decrease. Therefore, adding an impurity, will cause a substance to melt at a lower temperature.

In essence, melting is a process that is driven by the increase in entropy associated with going from the solid phase to the liquid phase. By making the liquid phase more disordered, this gain in entropy becomes larger, and melting becomes slightly more favorable.

Yes, I can see how an impurity added the liquid phase would increase the entropy, but for this to happen, the pure substance would have to begin melting at a depressed temperature in order to furnish some trace of liquid phase in which the impurity can dissolve. So we're kind of back to square one: why would a pure substance in contact with an impurity melt at a lower temperature.

The same applies to Raja's explanation. For that to work there would have to be liquid present. Where did this liquid phase come from if the mp of the pure substance hasn't been reached?

Or what would happen if substance and impurity were mutually insoluble? Then there should be no mp depression through entropy of mixing. (Should there?)
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Yggdrasil

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Re: Explanation of melting point depression in impure solids
« Reply #6 on: September 27, 2011, 05:14:27 AM »

The solid will be in dynamic equilibrium with the liquid, so that molecules at the edge of the solid will transiently move back and forth between solid-like and liquid-like phases.  Of course, at temperatures below the melting point, the equilibrium will lie far toward the solid phase, the solid can still sample the liquid phase.

There should be no mp depression if you add an insoluble substance to the liquid phase.
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