 Okay, so we've begun to understand this idea of freezing point depression. Why it is that both pure A and pure B, when I dissolve a little bit of the other in the first one, it will decrease the freezing point in both these directions. That has to do with the fact that the melting point is always lowered because the chemical potential is decreased in the solution relative to the pure liquid, so the intersection point with the curve for the solid has been shifted to lower temperatures. So our job now is to see if we can understand in a little more detail exactly how much shift in the temperature there's going to be for a certain amount of shift in a certain amount of solute dissolved in the solvent. So since we're talking about a phase equilibrium, equilibrium between the liquid phase and a solid phase, let's go back to what we understand about all phase equilibrium, is that when we're on this coexistence line, when we've just started to solidify the solid, the solid has the same chemical potential as the solution. A will be equally happy both in solution and in solid phase if the chemical potential is equal in those two phases. And we also know that the chemical potential of a liquid is related to the chemical potential of a solution is related to that of the liquid via this term that involves the activity of the solvent in that solution. So we have this expression. Since our goal is to understand something about how the activity relates to this temperature change, let's rearrange and isolate the activity. And let me also say that this temperature at which this is true, the temperature at which the chemical potential of the solid is equal to that of the solution, is the melting point, is the temperature at which this solid and solution are in equilibrium with each other. So if I rearrange this to solve for activity or the log of the activity, so first of all I need to get this chemical potential of a liquid over to this side. So chemical potential solid minus that of the liquid. And then I've left RT log A. So if I divide by RT, what I'm left with is log of activity. So now I've isolated the activity. The next step, we can actually, now that I've got chemical potential of the solid minus chemical potential of the liquid, notice that these are both single phases. This is not the solution anymore. This is the pure solid, chemical potential of A in a pure solid form. This is chemical potential of A in the pure liquid form. Remember also that chemical potential of A is the same as the partial molar Gibbs free energy of A. So the difference between the Gibbs free energy in the solid and liquid, that sounds a little bit like the Gibbs free energy of melting, the Gibbs free energy of fusion. That would be this quantity, the Gibbs free energy of fusion. Fusion always refers to solid melting, so the reactant is a solid, the product is a liquid, so that's for the process solid becoming liquid. The Gibbs free energy of fusion is free energy of products minus free energy of reactants, so that's partial molar Gibbs free energy of the liquid minus partial molar Gibbs free energy of the solid. And again, because partial molar Gibbs is the same as chemical potential, that's chemical potential of liquid minus chemical potential of solid. So that's what I had in mind when I said this sounds like the Gibbs free energy of fusion, but what we have up here in the numerator is chemical potential of A in the solid phase minus chemical potential of A in the liquid phase. If it were the other way around, if it were liquid minus solid, that would be exactly what we mean by the Gibbs free energy of fusion. This is the reverse of that. So I need to write it as negative Gibbs free energy of fusion of this quantity A divided by R, and let me write it divided by R and divided by T with this is a negative sign, so it's equal to negative one over R delta G of fusion over T. And the reason I've written it that way, because now G over T, that reminds me of something we've seen before, that reminds me of the Gibbs Helmholtz equation. So the Gibbs Helmholtz equation, what we're going to do next to this equation is we're going to take the temperature derivative. And again, the temperatures we're talking about here, I've forgotten to write temperature of fusion, but these are the melting temperatures that we're talking about. So I'm going to take the derivative with respect to this temperature, and Gibbs Helmholtz is going to remind me that when I take the temperature derivative of not G, but G over T, when I take the temperature derivative of G over T, what I get is negative H over T squared. That's the Gibbs Helmholtz equation. So that's what we're going to use in this next step. I need to take the derivative of both sides, and when I do that, I will get, so derivative of log A with respect to T is minus 1 over R, derivative of this thing with respect to T, derivative of G over T with respect to T is minus H over T squared. So I've got minus, that kills the negative sign I already had, H. So these are delta Gs of fusion. So this becomes a delta H of fusion. And now it's divided by T squared instead of a T. That's what I get after applying Gibbs Helmholtz. So that looks like something, right? Looks like I've made it a little more complicated at this point. But the good news is I've got this rearranged in a form that's going to be useful to us. So all I need to do now is if I break this derivative apart, derivative of log A with respect to T, if I move the DT over to the other side, so I'll write differential of log A is equal to 1 over R delta H of fusion over temperature melting point squared, multiplied by DT. That's now written in a form that we recognize what to do with this. I've got a differential on this side. I've got a DT on this side with some stuff that involves temperature dependent things before it. So I just need to throw an integral on both sides. So integrating log A, integrating 1 over T squared DT. And I guess I'm technically integrating with respect to this T fusion. I keep forgetting to write the subscript. But I'm integrating delta H over the melting point squared with respect to the melting point. The question now becomes that we have to think a little bit carefully about where am I integrating from and to? What are the limits of my integration? The answer to that question is I understand a lot about the melting point of the pure substance. I usually know what the melting point of the pure substance is. I want to know the melting point in the solution. I want to know what the melting point is after I've dissolved a little bit of solute into the solution. So we're going to integrate from the place we know something about over to the place we want to know about. So we're going to integrate from pure liquid to in the upper limit the solution in whatever concentration we're interested in. So when I do that for the pure liquid, we know for pure liquid the activity is 1 or the log of the activity would be 0. In the solution, the variable I'm integrating is not activity but log of activity. That's what's inside this differential. The value of this variable for the solution is just some particular value of log A. I don't know what that is yet. We'll talk about that a little bit later. But I'm integrating from the conditions where log of A is 0 because activity is 1 for pure solvent up to the point where log of A is some other value log A. Likewise over here I'm integrating from pure solution which has a freezing point. My integration variable is the freezing point. I'm integrating from the pure substances freezing point up to some new freezing point that I'm interested in. So the upper limit of integration here is T sub fusion. All right. So that's a very decorated equation. Let's do the integral and see what we get. Integral of d log A from here to here is just the upper limit log A minus the lower limit or 0. On the other side I've got a 1 over r. Integral of delta H over T squared. Here I have to stop and think and what we'll do is we'll make an assumption. I'm going to pull this delta H out of the integral. Certainly it's going to make my life easier if I do that. Enthalpes in general do depend on temperature. Heat capacities are what tell us how enthalpes depend on temperature. But what is going to happen here is I'm asking how much does that enthalpy of fusion change as I change the temperature by a little bit. Usually the freezing point is depressed by a few degrees. So as long as I'm not changing the freezing point by more than a couple degrees I'm going to assume the entropy, I'm sorry the enthalpy of fusion is not going to change by too much and I'll pull that out of the integral. So then all I have left is integral of 1 over T squared which is minus 1 over T evaluated between the pure freezing point and the freezing point I'm interested in. So now inserting the limits on the right side, subtracting 0 doesn't do anything. The left side so I've got log of activity is equal to a negative sign, enthalpy of fusion over R. I've already taken care of the negative sign so this difference is going to be 1 over the upper limit, 1 over the depressed freezing point, 1 over the freezing point of the solution minus 1 over the lower limit, 1 over the pure liquids freezing point. So we're almost done with the mathematical manipulations and then we'll stop and pause and make sure we understand what we've gotten. Let me go ahead and write this one more time simplifying instead of subtracting two fractions. So I'm going to write minus enthalpy of fusion over R and then I'll put these two fractions over the same denominator. So that denominator is going to look like one of these temperatures multiplied by the other temperature and then in order to get them over that denominator this one I needed to multiply by T0 and this one I needed to multiply by the T without the knot. So that's maybe not the final form. That's a significant intermediate result that we'll pause and talk about for just a second. What that tells us, that tells us a relationship between the activity of a solution. So I dissolve a little bit of solute in a solution and that lowers the activity from 1 to some lower value. If I know what that activity is now, this equation tells me the relationship between that activity and what the freezing point of the solution is. Not the original freezing point but the hopefully lower value of the freezing point that I've got by making that solution. Let's make sure we understand whether it's freezing point depression or not. So if I do prepare a solution that's not pure solvent, not pure A, but I've dissolved a little bit of solute in it, the activity of the solvent is going to go down. It's less than one. So the log of the activity, this quantity is going to be a negative number. So the left hand side is negative, the right hand side is also going to be negative. There's a negative sign here, enthalpies and R's and both of these temperatures in the denominator are positive numbers. So in order to make the right hand side negative with this negative sign I also need the numerator to be a negative number. So this quantity, I'm sorry I said that backwards, this negative sign I need this quantity I've just put in parentheses, I need that to be a positive number. So what I need to be true is for T0 to be bigger than T fusion. So when I take that difference I get a positive number. So I need the perturbed melting point to be lower than the original melting point. In other words the freezing point has been depressed, the freezing point has been lowered when I dissolved something in solution. So again qualitatively after doing all this math we've got something that makes sense. We're seeing freezing point depression because the chemical potential has been lowered but now we have this quantitative relationship telling us exactly how much the freezing point has been changed in response to a solution with a particular activity. So the next step and we'll save and do that in the next video is to make a relationship between the activity and the concentration of the solution. So that's coming up next.