Question

For a mixture of n-butane (1) + n-pentane(2) at 298.15 K, what would be the predicted bubble-point pressure using Raoult's Law if your mixture was 20% n-butane by mole Would you expect Raoult's law to be a good model for this system Why or why not

Answer 1

Raoult's Law only works for ideal mixtures.

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. An ideal mixture is one which obeys Raoult's Law, However, some liquid mixtures get fairly close to being ideal. These are mixtures of two very closely similar substances. like n-butane and n-pentane.

In 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 vapour 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. only half as many of each molecule escaping - but the proportion of each escaping is still the same. The diagram will be 50:50 mixture of the two liquids. It means that there are only half as many of each molecule on the surface as in the pure liquids.

The intermolecular attractions between two molecules of n-butane or n-pentane of each must all be exactly the same if the mixture is to be ideal. This is why mixtures get close to ideal behaviour. They are similarly sized molecules and so have similarly sized van der Waals attractions between them. However, they obviously aren't identical - and so although they get close to being ideal, they aren't actually ideal.

Raoult's Law

The definition below is the one to use if you are talking about mixtures of two volatile liquids.

"The partial vapour pressure of a component in a mixture is equal to the vapour pressure of the pure component at that temperature multiplies by its mole fraction in the mixture."

To predict possible vapour pressure at boiling point

For mixtures of n-butane and n-pentane, we expect that their boiling points would form a straight line joining the two points. we are going to add another line. This second line will show the composition of the vapour over the top of any particular boiling liquid.

If you boil a liquid mixture, you would expect to find that the more volatile substance escapes to form a vapour more easily than the less volatile one.

That means that in the case to find a higher proportion of n-butane (the more volatile component) in the vapour than in the liquid. we can discover this composition by condensing the vapour and analysing it. That would give us a point on the diagram.

The below diagram shows what happens if we boil a particular mixture. Notice that the vapour over the top of the boiling liquid has a composition which is much richer in n-butane - the more volatile component.

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