[Needaask]

for a positive deviation the explanation we learned was that the intermolecular forces of attraction between A-A and B-B individually are weaker than those between the A-B. Hence, the A and B repel each other allowing them to leave the solution to become a gas more easily. And the example given for this is a solution of carbon disulfide and acetone.

However, since carbon disulfide is non polar, it only has weak London Dispersion forces. So even though water-water intermolecular forces are greater than the water-carbon disulfide intermolecular forces, aren't the carbon disulfide intermolecular forces weaker than the water-carbon disulfide intermolecular forces?

If so, how is this an example of a solution with a positive deviation from Raoult's Law since only one of the individual components have a stronger intermolecular attraction than of the solution?

Thanks

[UG]

I do believe a solution of carbon disulphide and acetone means mixing carbon disulphide with acetone. Don't think there is any water involved.

[Needaask]

I do believe a solution of carbon disulphide and acetone means mixing carbon disulphide with acetone. Don't think there is any water involved.

Oops. I meant that the acetone-acetone interaction is stronger than the acetone-carbon disulfide interaction but definitely the carbon disulfide-carbon disulfide interactions is weaker than of the acetone-carbon disulfide. So in this case, how would the law work out?

[UG]

Basically you get a positive deviation if mixing is not energetically favourable in the liquid phase, this is usually the case for mixing polar and non-polar molecules. For some solutions, mixing is energetically favourable, such as ethanol-water mixtures, and you get a negative deviation

[Needaask]

Basically you get a positive deviation if mixing is not energetically favourable in the liquid phase, this is usually the case for mixing polar and non-polar molecules. For some solutions, mixing is energetically favourable, such as ethanol-water mixtures, and you get a negative deviation

But why is mixing non polar and polar stuff a positive deviation? Because now if A is the polar and B is the non polar component, A has stronger intermolecular interactions than AB But B has a weaker intermolecular interactions AB.

[UG]

To put it simply, the intermolecular force A-A is much greater than A-B which isn't much greater than B-B if that makes sense. Say you have some sort of arbitrary units for intermolecular force and you have 10 for A-A, 5 for A-B and 3 for B-B, then forming A-B is not energetically favourable, 10 + 3 > 2 x 5

[Needaask]

To put it simply, the intermolecular force A-A is much greater than A-B which isn't much greater than B-B if that makes sense. Say you have some sort of arbitrary units for intermolecular force and you have 10 for A-A, 5 for A-B and 3 for B-B, then forming A-B is not energetically favourable, 10 + 3 > 2 x 5

Ohh but in this case, isn't A-A>A-B>B-B?

Because according to my notes, A-A and B-B must be stronger than A-B for a positive deviation to occur. So I was thinking for this case, if we were to draw the phase diagram for A, it would be a positive deviation. Cos now there is more repulsion in A-B than in A-A so it's easier to get the A out of the solution.

But drawing the phase diagram of B, there would be a negative deviation? Cos now there is a greater attraction for A-B than in B-B? So because of that it would be harder for B to leave the solution?

Is this the right thought process?

[UG]

Because according to my notes, A-A and B-B must be stronger than A-B for a positive deviation to occur.

Hmm, well maybe this is the case (or is it A-A and B-B combined must be stronger?), but my gut feel is that there is no rule you can follow to predict whether a positive or negative deviation will occur, there are a lot of other factors that need to be considered.
For the next bit I am not sure what you mean, usually you would draw mole fraction of component A on the abscissa and vapour pressure on the ordinate, if there is a positive deviation, then both components will have larger vapour pressures than that calculated from Raoult's law.

[Needaask]

Because according to my notes, A-A and B-B must be stronger than A-B for a positive deviation to occur.

Hmm, well maybe this is the case (or is it A-A and B-B combined must be stronger?), but my gut feel is that there is no rule you can follow to predict whether a positive or negative deviation will occur, there are a lot of other factors that need to be considered.
For the next bit I am not sure what you mean, usually you would draw mole fraction of component A on the abscissa and vapour pressure on the ordinate, if there is a positive deviation, then both components will have larger vapour pressures than that calculated from Raoult's law.

Oh! I thought we should compare them individually? Not saying that the total intermolecular forces in A individually and the total intermolecular forces in B individually but be greater than the total forces in AB combined. Rather I was thinking that the average interactions in A individually and B individually then AB individually.

For the phase diagrams being drawn, using there would only be one graph with a positive deviation for Eg A. But is it possible for B to have a negative deviation?

[Needaask]

Oh wait, reading this link http://www.chemguide.co.uk/physical/phaseeqia/nonideal.html I see that the graph is for the total pressure and not the partial pressure of a gas but actually the pressure of both the gases!

So in a positive deviation, both gases are escaping at a faster rate? The explanation given was "That is because the intermolecular forces between molecules of A and B are less than they are in the pure liquids.
You can see this when you mix the liquids. Less heat is evolved when the new attractions are set up than was absorbed to break the original ones. Heat will therefore be absorbed when the liquids mix. The enthalpy change of mixing is endothermic."

So this would mean that the intermolecular strength between AB is stronger than in AA or BB individually? I'm a bit lost now on what their explanation here means.

Thanks

[UG]

See I would take that as the interaction of A and B molecules is less than interaction of A-A or B-B molecules, so perhaps your initial assumption that carbon disulfide-carbon disulfide interactions are weaker than of the acetone-carbon disulfide is incorrect.

[Needaask]

See I would take that as the interaction of A and B molecules is less than interaction of A-A or B-B molecules, so perhaps your initial assumption that carbon disulfide-carbon disulfide interactions are weaker than of the acetone-carbon disulfide is incorrect.

Ohh so the carbon disulfide interactions would still be greater than of the acetone-carbon disulfide interactions? Why is this so?

Thanks so much for the help

[UG]

Well the first thing that came to my mind was the very regular and tight packing of the linear CS₂ molecules, this arrangement would be destroyed with acetone in the mixture.

[Needaask]

Well the first thing that came to my mind was the very regular and tight packing of the linear CS₂ molecules, this arrangement would be destroyed with acetone in the mixture.

Hi thanks for the reply could you emphasize on this? The only thing I could think of was repulsion between the electrons of the two species. But I don't have any particular reason why the repulsion would be greater when the non polar and polar molecule come together actually.

Thanks

[UG]

For carbon disulphide there are only dispersion (instantaneous dipole–induced dipole) forces which is a weak intermolecular force, but the molecules are packed very tight together so the force only needs to act over a small distance. When the carbon disulphide interacts with acetone, you have some permanent dipole-induced dipole which is (theoretically) stronger, but the packing is not so well so the force has to act over more distance and so is weaker. Even permanent dipole-dipole could be weaker compared to dispersion forces if the molecules are chosen correctly, e.g. consider boiling point of I₂ which is 184°C (only dispersion forces) and HI (permanent dipole-dipole), boiling point -34°C