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I was reading about growing single crystals and I'm a little confused about this -

In most crystal growing processes, a "seed crystal" is used, and the rest of the material crystallizes on the seed crystal (for example the czochrlaski process).

But thinking about entropy, wouldn't it be more favourable for the vapour/liquid to not crystallize and hence maintain a larger disorder?

A similar process occurs in supersaturated vapour as well, where a small liquid droplet acts as a "seed" for more condensation to occur. So I suspect I'm missing something, but I can't figure out what.

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7 Answers 7

up vote 7 down vote accepted

Indeed this is a handy counter example to people who, not understanding the second law, claim that evolution is impossible on the basis that entropy decreases (nevermind that by this misunderstanding life itself would be impossible as well). The growing crystal is not a closed system: it exchanges energy and matter with the surrounding environment, and this can lead to a local entropy decrease if it is compensated by an entropy increase of the environment. The thermodynamic potential that is truly minimized, taking these exchanges into account, is the free energy. (Note that there are slightly different definitions of the free energy which apply to slightly different types of process, but they are all morally the same.)

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The nature of a crystallization process is governed by both thermodynamic and kinetic factors, which can make it highly variable and difficult to control. Factors such as impurity level, mixing regime, vessel design, and cooling profile can have a major impact on the size, number, and shape of crystals produced. Now put yourself in the place of a molecule within a pure and perfect crystal, being heated by an external source. At some sharply defined temperature, a bell rings, you must leave your neighbors, and the complicated architecture of the crystal collapses to that of a liquid. Textbook thermodynamics says that 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:

T(S liquid - S solid) > H liquid - H solid

and

G liquid < G solid

This rule suffers no exceptions when the temperature is rising. By the same token, on cooling the melt, at the very same temperature the bell should ring again, and molecules should click back into the very same crystalline form. The entropy decrease due to the ordering of molecules within the system is overcompensated by the thermal randomization of the surroundings, due to the release of the heat of fusion; the entropy of the universe increases. But liquids that behave in this way on cooling are the exception rather than the rule; in spite of the second principle of thermodynamics, crystallization usually occurs at lower temperatures (supercooling). This can only mean that a crystal is more easily destroyed than it is formed. Similarly, it is usually much easier to dissolve a perfect crystal in a solvent than to grow again a good crystal from the resulting solution. The nucleation and growth of a crystal are under kinetic, rather than thermodynamic, control.

Source

http://en.wikipedia.org/wiki/Crystallization

Also see (Nucleation and crystal growth, page 193. chapter 5: Kinetics, William M. White's book: Geochemistry, Wiley-Blackwell, 2013. UK)

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Broadly speaking the descriptions given are good, based on general arguments.

For some detail let us consider this:

The processes are important

One needs to make reference to the techniques and processes of crystal growth, in order to give an answer that reflects the phenomenon. The processes and conditions that the seed-vapour system are subjected to, during crystal growth, makes crystal growth an entirely different process to the one in which a sand castle loses its structural features over time, for example, from the point of view that crystal growth is not spontaneous in that sense.

The constant cooling of the seed-vapour system at constant crystal-growth-temperature $T_g$, at the seed end of the furnace, is telling us that there is heat loss at the constant temperature $T_g$, and this is associated with an entropy change

$\Delta S=-\frac{\Delta Q}{T_g}\lt 0$

This shows that the system is subjected to a process where its entropy decreases. Something similar to what happens with refrigerators. However, as said by others, this is compensated by an increase of the entropy of the room the furnace is in.

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Please don't ever let anyone tell you that entropy is disorder. This is one of those statements that's true for an ideal gas but not for much else. People often try to get around this by redefining "disorder" to mean "entropy", but that's just silly. It's best to think of entropy as what it is -- the logarithm of the number of microstates compatible with the system's measurable macroscopic state -- and not rely on the outdated "disorder" metaphor.

Having said that, the entropy of a crystal is actually lower than the entropy of the same substance in solution. This isn't paradoxical because, as others have pointed out, the crystallisation process releases heat, and this heat increases the temperature of the crystal's surroundings through the usual $dS = \delta Q/T$. Because of this, it's tempting to think that this means that the "disorder" created by heating the water outweighs the order created in forming the crystal.

But sometimes the heat that's given off can end up in the crystal itself rather than the surrounding water (this will typically happen with supercooled water becoming ice, for example) and in these cases one would end up having to say that a warm crystal is less ordered than cold liquid, which just doesn't really seem right. Better just to say that the system with the crystal in it has a higher entropy than the one without, even though, at least in the everyday sense, it has more order.

Some of the other answers have mentioned free energy, but this is really just a fancy way of saying that the total entropy of the crystal and its surrounding liquid has increased. The free energy is applied when the temperature of the whole system can be assumed to be constant. The change in total entropy is $\Delta S_\text{total} = \Delta S_\text{crystal} + \Delta S_\text{liquid}$ = $\Delta S_\text{crystal} + \Delta U_\text{liquid}/T$ = $\Delta S_\text{crystal} - \Delta U_\text{crystal}/T$, but because physicists are afraid of entropy they like to multiply this quantity by $-T$ for no particularly good reason and call it an energy instead: $\Delta A_\text{crystal} = -T\Delta S_\text{total} = \Delta U_\text{crystal} - T\Delta S_\text{crystal}$. You're allowed to multiply by $-T$ because it's assumed to be a constant. Because multiplying by $-T$ changes the sign we say that the free energy decreases, but this really just means that the total entropy increases.

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But as you have pointed out in another comment, isn't it possible that free energy decreases keeping entropy constant? As Colin McFaul pointed out in his answer, could it be that free energy is minimized even if the crystal's entropy decreases? –  Kitchi Feb 20 '13 at 17:57
    
@Kitchi the important thing to realise is that the free energy of the crystal decreases, but the entropy of the whole system increases. As long as the temperature remains constant, these are just two ways of saying the same thing. O In this particular case, as the solution turns into a crystal its entropy decreases, but it also loses energy (in the form of the heat of crystallisation, which is given off to the water)... –  Nathaniel Feb 21 '13 at 0:31
    
... You can either think of this as the crystal's free energy decreasing (because $U$ decreases faster than $TS$), or you can think of in terms of the water's entropy increasing because of the extra heat. So both of these things are true: the total entropy does increase, and also the crystal's free energy decreases even though its entropy decreases. –  Nathaniel Feb 21 '13 at 0:33
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@Nathaniel I think the statement “because physicists are afraid of entropy …” is somewhat out of context. On the contrary, physicists do like entropy, they have invested centuries studying it and trying to grasp its deeper meaning. But physicists also love energy, because at the end of the day, the bills are paid in terms of energy units. Just joking! –  JKL Feb 21 '13 at 1:31
    
@John yeah, I was kind of joking with that comment (and will happily edit it out if you don't think it's appropriate), but it's always seemed to me that the tradition of multiplying by $-T$ is a mistake. There are three reasons: firstly, it's only valid to make that step if $T$ is constant, whereas all you have to do to extend the concept to variable $T$ is to keep things in entropy units by not taking that step; –  Nathaniel Feb 21 '13 at 1:52

There are three things to consider.

  1. The system isn't necessarily going to maximize its entropy; it is going to minimize the Gibbs or Helmholtz free energy. These are not necessarily the same thing. The examples you gave are two good examples of this: the constituent particles are attracted to each other at some length scales, so that the energy of interactions is lower in the ordered state. At some temperature range, this energy "bonus" is great enough to overcome the entropy penalty. This is probably the biggest effect for the two cases you gave.

  2. The crystal is an open system, and can exchange energy and entropy with its environment. So it's not the free energy of the crystal that's being minimized; it's the free energy of the system (crystal + environment). Even if the free energy of the crystal is unfavorable for crystallization, it could still happen if the process increases the entropy of the environment enough. The best example I can come up with here is life, as @Michael Brown mentioned. Most biological processes work to keep the entropy in your body low, at the expense of increasing the entropy around you.

  3. Entropy sometimes behaves in ways that seem weird. For example, we usually think of mixing two substances as being entropically favored, even if there is no energy interaction. But carbon dioxide and water spontaneously demix because of the enormous entropy the gas gains by being freed of the water. This is almost certainly not relevant to either example you gave, but is a personal hobby horse of mine.

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I like the answer of Michael Brown, but it has a "hole". There is a simple counter-argument to this answer: do not consider the growing crystal alone, but take it with the whole environment: crystal, vapor and melt altogether. Then one again comes to the same question of Kitchi.

The origin of the misunderstanding is just the use of entropy as of a thermodynamic potential. Theoretically any potential might be used for description. In practice, however, the entropy is very inconvenient, since one of its variables is the mean internal energy. It has the effect that in order to be able to discuss a process in terms of its entropy one should make sure that the mean energy is constant during this process. Which is not the case during crystallization. Indeed, as soon as a molecule joins the crystal from the melt, the energy goes down.

The problem is removed, if one uses for these purposes the free energy, or even better the omega-potential, rather than the entropy. In the terms of these two potentials it is easier to think about phase transitions. After all, we got used to these terms much more, since they are akin to mechanical energy: the system goes to the state with the smallest free energy possible. In these same state the entropy will exhibit the absolute maximum.

One can now ask himself, if the free energy goes down during crystallization? The answer is "Yes, of course". Since, if under some conditions it does not go down with crystallization, the crystallization will not take place: only processes in which the free energy decreases leads to the thermodynamic equilibrium state. Then, does the total system entropy in this process go up? The answer is "Yes, of course", since, where the free energy goes down, the entropy necessarily goes up.

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"where the free energy goes down, the entropy necessarily goes up" - this is not necessarily true. $G=U-TS$, so the free energy can go down due to an increase in entropy or a decrease in internal energy, or even an increase in temperature while keeping the other two constant. –  Nathaniel Feb 20 '13 at 15:19

Probably, you are missing the latent heat impact on the entropy. The latent heat release increases the entropy of the liquid/vapor and overrides the decrease of entropy due to crystallization/condensation.

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Sorry, I just saw your answer go up as I was writing mine. –  Michael Brown Feb 20 '13 at 12:11
    
@MichaelBrown: It's OK, your answer is even better. –  Vladimir Kalitvianski Feb 20 '13 at 12:19

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