Polymorphism and Crystallization

[Theophrastus]

Hey guys,

I was thinking on the topics above the other day, and got an interesting question. Let's say you have a solution of a compound, which we'll call compound X. Compound X has two structural polymorphs; polymorph XY and polymorph XZ. Polymorph XZ much less stable (under the given conditions) as polymorph XY. (I know, a messed- up description, but bear with me...)

 

Now let's say you have a supersaturated solution of compound X. If you then rapidly cool the solution, adding a base crystal of XZ, will the solution, crystallizing around the base crystals, form polymorph XZ, or will the sample be ignored entirely, and eventually a X precipitate will allow a base for polymorph XY to grow?

 

ps: And just as a refresher, if two polymorphs are equally likely to form the process should work in providing preference of one over the other right?

[CaptainPanic]

I'm not sure that my post will actually answer the question. I've given it some thought, and I'm not 100% sure myself.

 

One problem is that you use poor wording, as you already knew: what do you mean by "stable"? If the crystal XZ is unstable, then it would form, and react to another shape crystal. If it's less favorable (which I'll describe below in more proper words), then... well... just read the following.

 

 

There are a number of factors that could influence the crystal growth, and I think these concepts overlap.

 

The parameter that describes whether a reaction will occur is the Gibbs Energy. There is plenty of info online (e.g. wikipedia). Basically, if the change of the Gibbs energy of reaction is negative, the reaction will proceed.

 

[math] \Delta G < 0 [/math], favored reaction (Spontaneous)

[math] \Delta G = 0 [/math], Neither the forward nor the reverse reaction prevails (Equilibrium)

[math] \Delta G > 0 [/math], disfavored reaction (Nonspontaneous)

 

Often, crystals will grow on a surface. This surface (you call it a "base crystal") will actually change the [math] \Delta G[/math]. So, in other words, if a certain type of crystal is already there, the type that will grow on it might be the same (since this is very often favored). the answer is therefore: "it depends...".

 

I am not sure if kinetics are also linked to the [math] \Delta G[/math]. I can imagine that with two equally favorable crystals don't necessarily have to form at the same speeds.

[Theophrastus]

One problem is that you use poor wording' date=' as you already knew: what do you mean by "stable"? If the crystal XZ is unstable, then it would form, and react to another shape crystal. If it's less favorable (which I'll describe below in more proper words), then... well... just read the following.

[/quote']

 

Thanks, I did intend to mean favorable, it's just that the definition of what is favorable for a compound can vary based on the conditions to which it is subjected. My use of the word stable came about from the idea that when a compound that has formed under given conditions, is subjected to different conditions in which, a different polymorph is favorable, it is seen as "unstable" under those conditions, as the crystal structure of the polymorph changes. For example, under S.T.P., the mineral aragonite (CaCO3) is unstable, "transforming" to calcite over time.

 

Thanks for the note on Gibbs. I already knew it before, but didn't think of the situation in the right context, in order to apply it. But as you hinted, the application of a crystal of the less favoured polymorph, could make our polymorph XZ more entropically favourable. Now that I think of it, this might make for a cool experiment, as in theory then, the reaction would be entropically favoured (my hypothesis, not necessarily fact, AFAIK) but enthalpically opposed, therefore polymorph XZ would only form when [math] T= /frac {/Delta H}{/Delta S} [/math], so by building a temperature range, one could find the ratio between [math] /Delta H [/math] and [math] /Delta S [/math], and then furthermore, knowing the standard dH at the given temperature, one can go further so as to find the precise change in entropy.

 

Right?

 

I am not sure if kinetics are also linked to the . I can imagine that with two equally favorable crystals don't necessarily have to form at the same speeds.

 

Well based on the conclusion I made above, and my earlier reference (I was refering more to preference of one over the other, that the rate of formation), the addition of a "base" crystal would increase the entropic favourability of one, resulting in a higher overall [math] /Delta G [/math], and thus favouring of its formation.

Edited September 19, 2009 by Theophrastus

[McCrunchy]

Hi Theo,

 

Let me first of all clarify what you meant by stable, less stable : you can define this by the solubility product Ks of the precipitation reaction. The lower the Ks, the more the ions will be prone to turn into a crystal, the more stable the crystal. Typically, since I guess you're talikng about CaCO3,

 

Ks-calcite < Ks-aragonite < Ks-vaterite

 

By the way, Ks is directly related to Delta G CaptainPanic mentionned by

Ks = exp(-Delta G/kT).

 

Ostwald's kinetic law states that the first polymorphs to form from a supersaturated solution are the most soluble ones, ie, that would be vaterite first. The reason for this is that the ions find their way to a crystal structure in steps : they first form loose, highly soluble structures with water inside - ikaite for example for CaCO3. Gradually, the water molecules are "squished" out of the crystal, but this requires overcoming energy barriers, and this takes time.

 

So thermodynamically the least soluble form is the most favoured one, but kinetically (Ostwald), its all reverse. The conclusions is that when kinetics are slow (low supersaturations), thermodynamics wins and you end up with calcite. On the contrary when you have high concentrations you typically get all sorts of polymorphs. To give you an example from my lab, mixing 20 mM NaHCO3 with 20 mM CaCL2 gives us a roughly a mixture half vaterite half calcite (determined from X-ray diffraction analysis).

 

I don't know how well Ostwald's empirical law applies; but man, it's a law, so it must be based on more than a couple observations!

 

Regarding your idea on seeding the solution with less stable polymorphs: I have indeed observed that seeds greatly reduce the induction time of precipitation reaction (ie, it accelerates the precipitation). This means that the ions accumulate on the seeds. Does this mean that they necessarily adopt the same polymorph ? Maybe. Only experiment can tell.

 

McCrunchy