Notes on Phase Seperation Chem 130A Fall 2002 Prof

Notes on Phase Seperation Chem 130A Fall 2002 Prof. Groves

Consider the process:

Pure 1 + Pure 2 Æ Mixture of 1 and 2

We have previously derived the entropy of mixing for this process from statistical principles and found it to be: ∆ =− − Smix kN11ln X kN 2 ln X 2

where N1 and N2 are the number of molecules of type 1 and 2, respectively. We use k, the

Boltzmann constant, since N1 and N2 are number of molecules. We could alternatively use R, the

gas constant, if we chose to represent the number of molecules in terms of moles. X1 and X2

represent the mole fraction (e.g. X1 = N1 / (N1+N2)) of 1 and 2, respectively.

If there are interactions between the two ∆ components, there will be a non-zero Hmix that we must consider. We can derive an expression ∆H = 2γ ∆ ∆ for Hmix using H of swapping one molecule of type 1, in pure type 1, with one molecule of type 2, in pure type 2 (see adjacent figure). We define γ as half of this ∆H of interchange. It is a differential interaction energy that tells us the energy difference between 1:1 and 2:2 interactions and 1:2 interactions. By defining 2γ = ∆H, γ represents the energy change associated with one molecule (note that two were involved in the swap).

∆ γ Now, to calculate Hmix from , we first consider 1 molecule of type 1 in the mixed system. For the moment we need not be concerned with whether or not they actually will mix, we just want ∆ to calculate Hmix as if they mix thoroughly. Then our result will allow us to decide if it can actually happen. Next, we ask what is the probability that the molecule adjacent to the one we

are considering is of type 2? The answer is X2. If half of the molecules are type 2, for example,

then X2 = 0.5 and the probability that the molecule adjacent to the one we are considering is 0.5.

This will, of course, be true for all N1 of the molecules of type 1. In our square representation

pictured above, each molecule has four immediate neighbors so there will be 4N1X2 of the 1:2 contacts in the well mixed system. This is the total number of 1:2 contacts. From above we

Copyright 2002 Jay T. Groves know that each 1:2 contact contributes γ/4, since there are 4 sides per molecule and we defined γ to correspond to one molecule. From this analysis, we find ∆ = γ HNXmix 12 =+γ ( ) NNXX1212

∆ ∆ ∆ Putting the pieces together, Gmix = –T Smix + Hmix, we find the free energy of mixing ∆ =+ ++γ ( ) Gmix kTN11ln X kTN 2 ln X 2 N 1212 N X X

Making some rearrangements using X1 + X2 = 1 and dividing though by (N1 + N2), we obtain an expression for the free energy on a per molecule basis: ∆ =+−( ) ( − ) +−γ ( ) GkTXXmix ( 11ln11 X 1 ln X 1) XX 1 1 1

This form is convenient ∆G γ because the free energy is mix = 3 kT a function of a single X1 variable, X1. By plotting ∆ G mix vs X1 for a variety γ of values of , we can see γ = 2kT the different behavior. Notice that for γ > 2kT, ∆ two minima in the Gmix γ = kT surface exist. These correspond to two coexisting phases. γ = 0

∆ Our expression for Gmix tells us what will happen when we attempt to mix components 1 and 2 γ based on the temperature, , and the molar ratio we mix together (e.g. X1). This information can be used to map out a phase diagram for the binary mixture. In the phase diagram, we will be interested in what happens to different composition mixtures at different temperatures so we will map X1 on one axis and T on the other. This is not the only way to create a phase diagram, but it will prove most convenient for the current application. For a particular system, the interaction energy, γ, is a constant. The value of γ (strength of the molecular interaction) sets the critical γ temperature, Tc, which is the highest temperature at which phase seperation can occur: = 2kTc.

01 X1

∆ The phase diagram pictured above was calculated from our expression for G mix. You can ∆ imagine that a different Gmix curve exists at each temperature, corresponding to horizontal lines ∆ traced across the diagram. If we plot points corresponding to the minima in the Gmix curves, these will trace the binodal and represent phase compositions that can coexist in equilbrium. If you attempt to prepare a mixure that falls inside this binodal curve, such as the light blue dot plotted on the graph, this mixutre will phase separate into two phases with compositions equal to the two corresponding compositions on the binodal curve at that temperature. For example the light blue mixture phase separates into white and dark blue as drawn. At higher temperatures, you can see that the phase compositions become more and more similar. Ultimately, at the critical temperature, Tc, and higher, only one fully mixed phase will exist.

Another thing we can explore with the free energy function is the mechanism by which phase ∂∂2∆ 2 separation occurs. If GXmix 1 is negative, then the mixture is intrinsically unstable and will rapidly separate into two phases. This occurs because fluctuations away from a homogeneous mixture actually lower the free energy and will thus be amplified. Think of it this way: you start with a patch of the mixture and a random fluctuation causes a few more of type 1 molecules to be on the left, leaving a small excess of type 2 on the right. The original mixture now really consists of two different mixtures, of composition X1+dX1 and X1-dX1. Now we ask whether the total free energy of these two new mixtures is more or less than that of the starting mixture. Referring to you calculus, you can convince yourself that the two new mixtures are ∂∂2∆ 2 lower in energy whenever GXmix 1 is negative and greater in energy whenever

Copyright 2002 Jay T. Groves ∂∂2∆ 2 ∆ GXmix 1 is positive. In the case of a positive second derivative of Gmix, the mixture is metastable. It will require nucleation for phase separation to occur. In general, nucleation does occur so you will always see phase separation below the binodal curve. The spinodal curve is ∂∂2∆ 2 defined as the locus of points where GXmix 1 = 0 and forms the boundary between intrinsically unstable and metastable mixtures.

This type of phase diagram can describe the behavior of cell membrane lipid mixutres quite accurately. Cell membranes generally exist in phase separated states and there is even some evidence indicating that cells actively maintain their membranes to be near a critical point for this type of demixing phase separation.

Extra effort problem: γ ≤ Prove: 2 kT implies homgeneous mixing will occur for all values of X1.

Hint: From phase diagram we have learned that highest temperature at which phase seperation ∆ can occur is the ciritical temperature, where both first and second derivatives of Gmix go to zero at the critical point. Compute the value of γ that is required for a mixture to reach criticality at T. (you should get 2kT)

Explain why mixtures with γ values below this can’t phase separate. ∆ Derivatives of Gmix are provided below for you to check your calculus.

∂∆G mix =−−kT(ln X ln(112 X)) +−γ ( X ) ∂ 11 1 X1

∂ 2 ∆G  11 mix =+kT  − 2γ ∂ 2  −  X1 XX111