 All right, so we've seen the freezing point depression, the fact that the freezing point of a solution is lower than the freezing point of the pure solvent. The presence of a solute solution will also affect the boiling point of that solution. So the next step is to understand how that works. So from a qualitative point of view, the reason we have a freezing point depression is if we plot the chemical potential of a solid liquid or gas versus temperature, they look like this. In a solution, the chemical potential is lowered. So the lower chemical potential in a solution compared to the liquid means that the freezing point got shifted to a lower temperature. The freezing point got depressed. Likewise, for exactly the same reason, the chemical potential of the liquid got lowered. So that shifts the boiling point of the solution over to a higher temperature. So while we have freezing point depression, we end up with boiling point elevation. The equation here reminds us what we know about freezing point depression, proportional to the log of the activity of the solvent, or if we assume dilute solution and make some approximations, proportional to the molality of the solute. The question then is what can we determine likewise about the change in the boiling point of the solution? And the derivation is going to be essentially exactly the same as the freezing point depression, derivation for the freezing point depression. Clearly, since one of them, the freezing point gets depressed and the boiling point gets elevated, there's going to be a sign difference that pops up somewhere. So the math itself is not all that useful. I'm going to go relatively quickly through the derivation. You can perhaps use this as a check to make sure you understood the steps in the freezing point depression to make sure you understand the steps in this derivation as well. But the key thing to watch for is going to be where the sign difference comes in that results in a boiling point elevation as opposed to a freezing point depression. So we're going to start in the same way if we have a solution of some solvent A and a solute B. Except now, instead of talking about freezing that solution, we're going to talk about boiling that solution. So we have equilibrium between the solution phase and the gas phase. The chemical potential, if we're at equilibrium, is equal in the gas phase and in the solution phase. And the chemical potential of the solution itself is related to the chemical potential of the liquid with an additional term involving the activity of the solvent in that particular solution. So if I rearrange this equation, chemical potential of the gas relative to the liquid with this log activity term, if I rearrange that to solve for the activity, I'll end up with chemical potential of the gas minus that of the liquid over RT. But if we remember chemical potential, that's just partial molar Gibbs free energy. So the numerator looks like partial molar Gibbs free energy of a gas minus partial molar Gibbs free energy of the liquid. And since those are pure phases, molar energy and partial molar energy are the same thing for the pure gas and the pure liquid. So the difference of Gibbs free energy for a gas minus Gibbs free energy of the liquid, that's exactly the vaporization Gibbs free energy. If I have the reaction of liquid boiling and converting into gas, the reactant is the liquid, the product is the gas. So this is Gibbs free energy of the product minus Gibbs free energy of the reactant. So that numerator I can write as the Gibbs free energy of the reaction, not just of any reaction, that particular reaction is the vaporization reaction. So the Gibbs free energy of vaporization divided by RT. And I suppose it's worth pausing a minute to make sure we understand what T we're talking about here. This is the temperature at which the liquid, the solution, is in equilibrium with its vapor. So the temperature at which the solution is boiling. So this temperature that we're talking about is the vaporization temperature in all of these equations. And as we'll see in just a second, that vaporization temperature is not the same as the vaporization temperature of the boiling point for the pure liquid. So this is the vaporization temperature of the boiling point of the solution. All right, so next step is to use the Gibbs-Helmholtz equation. If I take the derivative of log A with respect to the temperature, derivative of delta G over T with respect to temperature, according to Gibbs-Helmholtz is negative derivative of delta H over T squared. Now what I do with this expression, and here's where I'll skip a few steps worth of the math, is to separate the variables, get the temperatures over the right side, integrate that expression. So when I integrate on the left side, I'll get log of activity. On the right side, integral of minus 1 over T squared becomes positive 1 over T. And if I do that in between the pure solvent and the solution, I'm going to have delta H over T. Sorry, I've got R that I've left out. Delta G over RT, when I integrate, becomes RT squared. Delta H over R, the 1 over T, I'll write as 1 over T minus 1 over T naught. Or I can rewrite that as delta H over R. If I combine these into one fraction, it's going to look like T naught, boiling point of the pure solvent, minus T, boiling point of the solution. And then in the denominator, I have a product of those two, the boiling point and the pure solvent boiling point. Next step is I can do two things. Number one, I can define this temperature difference. So the boiling point elevation we're interested in. We've convinced ourselves that the boiling point in the solution is going to be a higher number than the boiling point, the lower boiling point of the pure solvent. So that amount of elevation of the boiling point is going to be by how much the boiling point has increased relative to the boiling point of the pure solvent. So with that definition, I've defined it this way because I know the boiling point will be higher than the pure solvent boiling point. This numerator, therefore, is going to be negative. The boiling point of the pure solvent is a smaller number than the boiling point of the solution. So this difference is negative. That makes sense because the activity of the solvent is less than 1 and this logarithm is going to be negative. So both sides of this equation are negative. So I'll write down negative delta t in the numerator. This difference in the numerator is negative delta t in the denominator. Rather than writing boiling point times pure solvent boiling point, I'll make the same approximation we did for freezing point depression and say in a relatively dilute solution, those two numbers are very close to each other in order to be able to solve easily for the boiling point. I'll just approximate it in the denominator of this expression by the boiling point for the pure solvent. So we have this expression. Log of the activity is negative enthalpy of vaporization divided by the gas constant times boiling point elevation divided by the pure solvent boiling point squared. That is the equivalent of this expression. In fact, if I rearrange this equation algebraically to solve for the boiling point elevation itself, that's going to look like minus RT squared, boiling point squared divided by the enthalpy of vaporization multiplied by log of the activity. And that expression looks just like this expression. Note a couple of things. There's no sign difference. The amount of the elevation that we've defined here has a negative sign in these constants. The amount of the freezing point depression also has a negative sign in the exact same constants, except it's freezing point and enthalpy of fusion rather than boiling point and enthalpy of vaporization. The sign difference I mentioned, boiling point elevation, freezing point depression comes because of our definition of what they mean. This is the boiling point elevation. We defined the freezing point depression assuming already that the freezing point is decreasing. So the sign difference comes in the definition of these quantities. In this derivation, the place where the sign change originated is right here. In the derivation of the boiling point elevation, our delta G of vaporization involved products minus reactants, gas minus liquid. When we were talking about freezing point, when we're talking about a fusion process, a melting process, that's solid converting to liquid. So the difference there was products or liquid minus the reactants which are the solid. So in this case, the liquid is the product. In this case, the liquid is the reactant. And so the sign of this numerator was reversed in these two cases. So that was the origin of the sign change mathematically that led to boiling point elevation rather than freezing point depression. So we have this relatively accurate expression for the boiling point elevation. We can do the same thing that we did for freezing point depression and make a few approximations in a dilute solution where the log of the activity is similar to the log of the mole fraction. And that's mole fraction of solvent, which is the same as 1 minus the mole fraction of solute. This ends up being, again, using the same steps we used for freezing point depression after approximating that log of 1 minus mole fraction of solute with its Taylor series, we can say this is at least approximately equal to in dilute solution this collection of constants times the concentration of the solute in units of mole fraction. And if we'd rather write that in units of molarity rather than mole fraction, that ends up introducing an additional molar mass of the solvent. That should be not melting point, but boiling point of the solvent squared denominator has enthalpy of vaporization. So again, that's making some additional approximations to replace mole fraction with molality. And lastly, I'm sorry, this negative sign should not be there. Log of activity becomes negative mole fraction of solute that absorbs this negative sign. So the negative signs are not here. This quantity, then, that I'll put in parentheses is what we call the boiling point elevation constant. So the term I put in parentheses is labeled k sub b. So we can compute that boiling point elevation constant. Again, just like with freezing point depression, that boiling point elevation constant is different for every different solvent because every solvent has a different enthalpy of vaporization and a different boiling point and a different molar mass. But if we combine those constants, we can calculate this boiling point elevation constant and use that to calculate the magnitude of a boiling point elevation for a solution with a particular concentration. So that's what we'll do next.