Q:

Entropy and Crystal Growth

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.

A:

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.)

A:

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logarithm of the number of microstates compatible with the system's measurable macroscopic state -- and not rely on the outdated "disorder" metaphor.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 $\Delta S_\text{total} = \Delta S_\text{crystal} + \Delta S_\text{liquid}$ =  $\Delta S_\text{crystal} + \Delta U_\text{liquid}/T$ = $\Delta S_\text{crystal} - \Delta U_\text{crystal}/T$. However, there is a tradition in physics to multiply this by $-T$, which puts it into energy units: $\Delta A_\text{crystal} = -T\Delta S_\text{total} = \Delta U_\text{crystal} - T\Delta S_\text{crystal}$. You're allowed to multiply by $-T$ because it's assumed to be a constant. Note, though, that although this puts it into energy units, we're still really dealing with entropy. Because multiplying by $-T$ changes the sign we say that the free energy decreases, but this really just means that the total entropy increases.

