Confusion over melting & crystallization?

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Main Question or Discussion Point

Say we had some solid and it's being heated up towards it's melting point. It reaches that temperature, and then an infinitesimal increase in temperature T will cause the crystal to break apart. It loses its ordered structure and so the entropy S has increased.
If I'm not mistaken, any spontaneous process occurs if it will allow a system to lower its free energy G, ie. reduce the useful energy it has available to do work, where it is defined as
G = U + pV - TS = H - TS, where enthalpy H = U + pV.
similarly the change in free energy is [tex] \Delta G = \Delta U + p\Delta V - T\Delta S [/tex].

So, when the entropy change [tex] \Delta S [/tex] is positive, the [tex] -T\Delta S [/tex] term is negative. That means if [tex]|-T\Delta S| > |\Delta H| [/tex] the change in free energy [tex] \Delta G [/tex] will be negative and melting will indeed happen, right?

My first issue is with this enthalpy term, H. I'm stuggling to picture how it will change. First we have the internal energy U, which is the Coulomb interaction between neighbouring molecules, right? I'm imagining this as an attractive potential, so when the solid melts it increases the separation of its molecules, which must mean an increase in potential energy. Thus U increases, which is a positive contribution to H.

Then we have the pV term. This is the mechanical work a system does on its environment, right? So by melting (if we aren't dealing with water), the volume of the stuff that was initially in the solid state (ie. our "system"?) has increased, which is a further positive increase to [tex] \Delta H [/tex]. It has expanded out, exerting a force on the fluid initially surrounding it (its "environment"?) So, if I'm correct about everything so far, melting involves an increase in enthalpy, right? But it only proceeds if the free energy loss of the [tex] -T\Delta S [/tex] term outweighs it?
This is the opposite of what I keep reading. For example, on wikipedia, "...melting occurs because the entropy, S, gain in your system by spatial randomization of the molecules has overcome the enthalpy, H, loss due to breaking the crystal packing forces."

So what's going on here, where am I getting mixed up?


I'm also confused over what happens when going the other way, cooling a liquid down to its melting point, which is defined as the point where the free energies of liquid & solid are the same. So, if we started forming a solid from a liquid, we have decreased the entropy of the molecules involved as they have taken on a more ordered arrangement. This negative increase in entropy would give a positive [tex] \Delta G[/tex], so why should this process ever occur?
 
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Answers and Replies

  • #2
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You are nearly there all that needs doing is a little tidying of your argument.

First there is no 'infinitesimal temperature change'

Think back to cooling curves - you must have studied these before coming to full blown thermo.

So long as there is some solid and some liquid there is no temperature change. The two states are in equilibrium so the process is thermodynamically reversible (an important point as it allows us to use the equations you quote).

Although there is no temperature change heat is evolved by fusion or taken up by melting and is known as latent heat.

In the case of melting the volume change is negligable, compared to this, so the P[tex]\Delta[/tex]V work can be disregarded (you can account for it if you like)

So the change in enthalphy

[tex]\Delta[/tex]H = q

and [tex]\Delta[/tex]H = T[tex]\Delta[/tex]S

So at this stage of a cooling curve

[tex]\Delta[/tex]G = 0

and the process is reversible.

To summarise

Melting latent heat gain ([tex]\Delta[/tex]H +ve) entropy gain (minus T[tex]\Delta[/tex]S -ve)

Solidifying latent heat loss ([tex]\Delta[/tex]H -ve) entropy loss (minus T[tex]\Delta[/tex]S +ve)


Hope this helps, go well.
 
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  • #3
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so then, in the question of why materials crystallize, we can say that the entropic free energy gain is outweighed by the enthalpy loss?
What about the extra interfacial energy that arises from forming a solid, where does that come in?
 
  • #4
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we can say that the entropic free energy gain is outweighed by the enthalpy loss?
No I said it exactly balances.

[tex]\Delta[/tex]H = H(liquid) - H (solid) = latent heat added to melt substance = qreversible

This is positive since heat is added.

Similarly

[tex]\Delta[/tex]S = S(liquid) - S (solid) = [tex]\Delta[/tex]H / T = qreversible / T

This is called the entropy of melting.
It is positive.

Take one away from the other and [tex]\Delta[/tex]G is zero.

Are you studying free energy from a physics or chemistry point of view?

Physicists tend to look for systems in equilibrium to do their thermodynamics on. By definition [tex]\Delta[/tex]G is zero in equilibrium.

Chemists tend to want reactions to proceed to a conclusion so look for non equilibrium situations where [tex]\Delta[/tex]G is nonzero
 
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  • #5
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I'm a physics student so I'm approaching this from a physics angle.

If it exactly balances, then why should any phase transition happen at all? I thought you needed a negative free energy change to make a process happen?
 
  • #6
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The strict condition is [tex]\Delta[/tex]G less than or equal to zero for a process to be possible.

leaving

[tex]\Delta[/tex]G greater than zero for mission impossible.

Incidentally this condition does not 'make things happen'.

It says nothing whatsoever about the rate the process proceeds at, so [tex]\Delta[/tex]G may be well negative but the process so slow as to be regarded as stopped.

An example of this would be the settlement of colloidal suspensions under gravity. This is (free) energy favourable, but there are jars of stuff, held by the Royal Society, which have not settled in nearly two centuries.
 
  • #7
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There is one other point I should have made.

The melting system described is not an isolated system. Heat (the latent heat) is added to the system from the rest of the universe or melting would not occur.
 
  • #8
Mapes
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What about the extra interfacial energy that arises from forming a solid, where does that come in?
In practice, this effect reduces the freezing temperature to some level below the melting temperature because of the energy cost of making a new surface. If the cooling rate is small and there are plenty of defects on the container's surface where the new solid can nucleate, the reduction is minimized.
 
  • #9
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hmm, well, I'm now wondering about if we had a polymer lamella crystal in a melt, and we were adding a single strand to the lamella. If we were doing this at the melting point of the lamella, this should involve a free energy change of zero, since the melting point is when liquid & solid phases have equal free energies, right?

well adding a single stem must increase the interfacial area on the top & bottom of the lamellar structure, that much is straightforward.
However, the expression I have for this free energy change is
[tex] \Delta G = 2a^2\sigma - \Delta H_m\frac{\Delta T}{T_m(\infty)} la^2 = 0[/tex]
where the undercooling you mention is [tex] \Delta T = T_m(\infty) - T [/tex]
and here [tex] T = T_m(l) [/tex] where the Tm are melting temperatures as a function of lamellar length l, Hm is the latent heat of melting/fusion and a is the length of one monomere in the polymer.

the first term is for the interfacial energy, but I'm struggling to picture what's happening in the second term.
Why is the free energy change of adding a stem to the crystal zero at the melting point? What makes it so that the second term compensates for the first?
 
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