0.17 Phase equilibrium in solutions  (Page 4/9)

Our starting point will be the dynamic equilibrium model we developed in the previous Concept Development Studies. Remember the essential point of this model: at liquid-vapor equilibrium, the rate of evaporation of the liquid is exactly equal to the rate of the condensation of the vapor. As we have just observed, the presence of a solute disrupts this equilibrium: adding a nonvolatile solute to a boiling solvent stops the boiling until the temperature is raised. Our question is then, which of the processes was disrupted: condensation, evaporation, or both?

Let’s start with condensation of the vapor. Remember that the solute is nonvolatile, which means that there are no solute molecules in the vapor phase. This indicates that the rate of condensation depends only on the number of moles per volume of solvent molecules in the vapor phase. When we add solute to the liquid phase solution, that number does not change. So the rate of condensation is the same, whether the solute is present or not, when the pressure of the vapor is the same.

This means that it is the rate of evaporation that must be disrupted by the presence of the solute. That does make sense, since we added the solute only to the liquid phase and evaporation is the process by which molecules leave the liquid phase and enter the vapor phase. Furthermore, after adding the solute, we observed that a new equilibrium is established at a lower vapor pressure. This means that at equilibrium the rate of condensation is lower than it was before we added the solute. So, at equilibrium, the rate of evaporation must be lower once we add the solute.

We can conclude that the rate of evaporation is lowered by the presence of a nonvolatile solute. This means that, for dynamic equilibrium, the rate of condensation must also be lower, which means that the pressure of the vapor must be lower at equilibrium.

What remains is to work out why the rate of evaporation is lowered by the solute. Here it is helpful to remember the second observation of Raoult’s law, which concludes that the amount of the vapor pressure lowering depends only on what fraction of the moles of particles present in the solution are solute particles. This is an important clue. Evaporation requires that solvent molecules overcome intermolecular attractions and “escape” the liquid phase. Not every solvent molecule is well positioned for an escape, regardless of whether it has sufficient kinetic energy. A solvent molecule must be near the surface of the liquid if it is to escape unfettered by running into other solvent molecules that will slow it down. Since the rate of evaporation is lower when the solute is present, the solute particles must reduce the number of solvent molecules which are positioned and available for evaporation near the surface of the liquid. The solute molecules impair the ability of solvent molecules to escape. And they do so in ways which do not depend on what the solute molecules are.

Why would the solute molecules impair the escape of the solvent molecules into the vapor? There are two ways to think about this, one more complicated than the other. A simple picture would be that, on and near the surface of the liquid, some of the surface area of the solution is occupied by the solute particles. The fraction of that surface taken up by solute molecules could be assumed to be the same proportion as the fraction of solute molecules in the solution. This would immediately account for Raoult’s law: solute molecules near the surface reduce the rate of evaporation in proportion to their mole fraction by taking up space near the surface. To match this lower rate of evaporation, the vapor pressure must be lower so that the rate of condensation is lower to achieve equilibrium.