This page deals with Raoult"s Law and how it applies to mixtures of two volatile liquids. It covers cases where the two liquids are entirely miscible in all proportions to give a single liquid - NOT those where one liquid floats on top of the other (immiscible liquids). The page explains what is meant by an ideal mixture and looks at how the phase diagram for such a mixture is built up and used.

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## Ideal Mixtures

An ideal mixture is one which obeys Raoult"s Law, but I want to look at the characteristics of an ideal mixture before actually stating Raoult"s Law. The page will flow better if I do it this way around. There is actually no such thing as an ideal mixture! However, some liquid mixtures get fairly close to being ideal. These are mixtures of two very closely similar substances. Commonly quoted examples include:

hexane and heptane benzene and methylbenzene propan-1-ol and propan-2-olIn a pure liquid, some of the more energetic molecules have enough energy to overcome the intermolecular attractions and escape from the surface to form a vapor. The smaller the intermolecular forces, the more molecules will be able to escape at any particular temperature.

If you have a second liquid, the same thing is true. At any particular temperature a certain proportion of the molecules will have enough energy to leave the surface.

In an ideal mixture of these two liquids, the tendency of the two different sorts of molecules to escape is unchanged.

You might think that the diagram shows only half as many of each molecule escaping - but the proportion of each escaping is still the same. The diagram is for a 50/50 mixture of the two liquids. That means that there are only half as many of each sort of molecule on the surface as in the pure liquids. If the proportion of each escaping stays the same, obviously only half as many will escape in any given time. If the red molecules still have the same tendency to escape as before, that must mean that the intermolecular forces between two red molecules must be exactly the same as the intermolecular forces between a red and a blue molecule.

If the forces were any different, the tendency to escape would change. Exactly the same thing is true of the forces between two blue molecules and the forces between a blue and a red. They must also be the same otherwise the blue ones would have a different tendency to escape than before. If you follow the logic of this through, the intermolecular attractions between two red molecules, two blue molecules or a red and a blue molecule must all be exactly the same if the mixture is to be ideal.

This is why mixtures like hexane and heptane get close to ideal behavior. They are similarly sized molecules and so have similarly sized van der Waals attractions between them. However, they obviously are not identical - and so although they get close to being ideal, they are not actually ideal. For the purposes of this topic, getting close to ideal is good enough!

Raoult"s Law only works for ideal mixtures. In equation form, for a mixture of liquids A and B, this reads:

\< P_A = \chi_A P^o_A \label{1}\>

\< P_B = \chi_B P^o_B \label{2}\>

In this equation, **PA** and **PB** are the **partial vapor pressures** of the components A and B. In any mixture of gases, each gas exerts its own pressure. This is called its partial pressure and is independent of the other gases present. Even if you took all the other gases away, the remaining gas would still be exerting its own partial pressure. The total vapor pressure of the mixture is equal to the sum of the individual partial pressures.

\< \underset{\text{total vapor pressure}}{P_{total} } = P_A + P_B \label{3}\>

The Po values are the vapor pressures of A and B if they were on their own as pure liquids. xA and xB are the mole fractions of A and B. That is exactly what it says it is - the fraction of the total number of moles present which is A or B. You calculate mole fraction using, for example:

\< \chi_A = \dfrac{\text{moles of A}}{\text{total number of moles}} \label{4}\>

## Vapor Pressure and Composition Diagrams

Suppose you have an ideal mixture of two liquids A and B. Each of A and B is making its own contribution to the overall vapor pressure of the mixture - as we"ve seen above. Let"s focus on one of these liquids - A, for example. Suppose you double the mole fraction of A in the mixture (keeping the temperature constant). According to Raoult"s Law, you will double its partial vapor pressure. If you triple the mole fraction, its partial vapor pressure will triple - and so on. In other words, the partial vapor pressure of A at a particular temperature is proportional to its mole fraction. If you plot a graph of the partial vapor pressure of A against its mole fraction, you will get a straight line.

Now we"ll do the same thing for B - except that we will plot it on the same set of axes. The mole fraction of B falls as A increases so the line will slope down rather than up. As the mole fraction of B falls, its vapor pressure will fall at the same rate.

Notice that the vapor pressure of pure B is higher than that of pure A. That means that molecules must break away more easily from the surface of B than of A. B is the more volatile liquid. To get the total vapor pressure of the mixture, you need to add the values for A and B together at each composition. The net effect of that is to give you a straight line as shown in the next diagram.

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We"re going to convert this into a boiling point / composition diagram. We"ll start with the boiling points of pure A and B. Since B has the higher vapor pressure, it will have the lower boiling point. If that is not obvious to you, go back and read the last section again!

For mixtures of A and B, you might perhaps have expected that their boiling points would form a straight line joining the two points we"ve already got. Not so! In fact, it turns out to be a curve.