Jesse

True, but again, if the system is in isolation entropy is guaranteed to increase in any spontaneous reaction, and oil and water will separate spontaneously even when totally isolated. As this page points out, the requirement that free energy decreases in a spontaneous reaction is actually equivalent to the requirement that the entropy of the molecules involved in the reaction + the entropy of their surroundings must always increase in a spontaneous reaction (any heat released in the reaction increases the entropy of the surroundings). If you use the Gibbs equation dG=dH-TdS, you can divide both sides by -T to get -dG/T=-dH/T+dS, which is equivalent to dS(universe)=dS(surroundings)+dS(molecules), as explained on the page above. So, if dG must always be negative in a spontaneous reaction, -dG/T = dS(universe) must always be positive. In some spontaneous reactions, even though the dS(molecules) is negative, the total entropy change indicated by -dG/T is still positive.

In other words, for a system in complete isolation in which a spontaneous chemical reaction occurs, the change in free energy is just the change in entropy of the system as a whole, multiplied by -T.

Supramolecular

Yes, and the oil/water system behaves in this way (i.e. free energy change is negative due to a +ve dS and a zero dH. The process would be even more favourable if there was an exothermic component to the process. As you mention, the heat released would increase the entropy of the surroundings.

I think we're singing from the same hymn-sheet as it were.

Regards

Goober

I think you guys have worked it out, but I'll post this anyway to clarify.

Seperate oil and water layers have less entropy than and oil-water mixture.

BUT, to form an oil water mixture would require energy. Why? Because water molecules are polar, and are held together by intermolecular bonds between polar regions. Oil is non-polar, so it will not form dipole-dipole bonds with water. This means to form an oil water mixture, you would need to break many of these dipole-dipole bonds to fit the oil molecules in. To do this, energy is required.

Where does this energy come from? It would come from heat, lowering the temperature of the water, oil and it's surroundings marginally. But this is an entropy decrease. This entropy decrease is larger than the increase in entropy that would be generated by them mixing, so the total entropy change is negative.

So it will not occur.

Edited to add: I re-read this, and noticed some of you are using the assumption that the enthalpy change is zero. But this neglects the fact that energy would be required to break the IM bonds between water molecules to make spaces for the oil molecules to occupy. Since oil has no strong IM interations with water, this energy would not be regained, hence DH is positive in my opinion. DS is positive in my opinion, but DH>T*DS, so DG>0.

Jesse

Goober:
I think you guys have worked it out, but I'll post this anyway to clarify.

Seperate oil and water layers have less entropy than and oil-water mixture.

Less total entropy? If that were true, then oil and water in isolation would become mixed, because the 2nd law says the entropy of an isolated system will always increase until it reaches a state of maximum entropy. My understanding was that separated oil and water layers actually have more entropy, because the rotational freedom of the water molecules is lowered when they're next to oil molecules, so minimizing the surface area of contact between oil and water gives the maximum number of water molecules the freedom to rotate. There will be some kind of trade-off between the rotational and translational (position of center of mass) degrees of freedom here, since there are fewer ways to arrange the centers of mass where they're separated than when they're mixed, but my understanding was that at maximum entropy the oil and water will still be mostly separated.

There may also be a difference in potential energy between the two states, in which case the state with lower potential would have greater kinetic energy in the form of heat, which would give it more possible microstates in the momentum phase space as well. This would also be part of the "trade-off". I'm not sure about this one though, since Supramolecular claimed dH is zero for oil and water becoming separated, while you claim below that it would be positive.

Goober:
BUT, to form an oil water mixture would require energy. Why? Because water molecules are polar, and are held together by intermolecular bonds between polar regions. Oil is non-polar, so it will not form dipole-dipole bonds with water. This means to form an oil water mixture, you would need to break many of these dipole-dipole bonds to fit the oil molecules in. To do this, energy is required.

Where does this energy come from? It would come from heat, lowering the temperature of the water, oil and it's surroundings marginally. But this is an entropy decrease. This entropy decrease is larger than the increase in entropy that would be generated by them mixing, so the total entropy change is negative.

But then you're contradicting your earlier statement that the mixed state has a greater total entropy than the separated state, no? Were you not talking about the total entropy of the system, but just about the dS in the Gibbs equation? Remember, dS != total entropy change, since any heat absorbed or released in the dH term is also part of the entropy change of the whole system.

Supramolecular

Goober,

The enthalpy change in this system is slightly +ve but is far offset by the larger +ve entropy change, which as both Jesse and I have mentioned is caused by the release of ordered water from around the oil molecules (i.e. water with less freedom to rotate than in bulk water). When the surface area of the oil molecules is decreased by coming together, water molecules are released back into bulk water. This results in a -ve dG term, hence the process occurs spontaneously. You statement that dG is +ve is incorrect. If this was the case, oil and water wouldn't seperate.

Your statement that "Seperate oil and water layers have less entropy than and oil-water mixture" is incorrect for the reasons Jesse mentioned and I stated above.

You must also remember the attractive Van Der Waals interactions between the oil molecules, which are stronger between oil-oil than oil-water.

Here's the equation below (obviously with made-up numbers).

dG = dH-TdS

dH = +1 (i.e. small & +ve), TdS = +20 (large & +ve)

Therefore dG = +1 - (+20) = -19, which is negative and so the process occurs spontaneously.

I wish I had my PhD thesis in PC format as I have a great figure explaining the hydrophobic effect in the introduction.
Some of this is quite counterintuitive e.g. it's easy to think that a mixture of oil and water is more disordered than two separate layers and so has a higher entropy but this would be wrong.

Also the study of non-molecular interactions is really very complicated by having to take a lot of competing interactions into acccount in a given process.
We still don't know how strong these non-covalent interactions are, even hydrogen bonds (read some of Dudley William's papers on vancomycin and the measurents of the strength of the hydrogen bond). Add to this weaker interactions such as pi-pi interactions that although are weaker can be numerous and co-operative. This is why it is so hard to predict the tertiary structure of proteins, or predict the single crystal structure of a compound.

I myself measured the strength of several non-covalent interactions including a pyrrole(NH)-pi interaction using synthesised molecular zippers. This type of interaction is common in various proteins and other crystal structures. It's strength in chloroform was 5 kJ/mol, far higher than we had thought and not far off from a hydrogen bond.

Actually, does anyone know how I could get my thesis from Mac format (circa 1995) floppy disks onto my PC. It's written using word. I like to be able to do it as a backup as I'd hate to lose my printed hardback copy? Would I lose a lot of word formatting in text and figures?

Cheers

Goober

I was talking about the process of mixing, not the process of seperation. DG is +ve for this process. Since seperation is just the reverse of mixing, DG +ve for mixing implyies DG is -ve for seperation. I did it this way because it is easier to consider mixing than seperation. Could we possibly discuss it from this perspective, I honestly think it is easier this way.

Yes I only considered dipole bonds, but this only makes DH more positive. You are breaking both dispersion bonds and dipole bonds and replacing them with virtually nothing. Hence and energy input is required, hence DH is +ve.

IMO, the water around oil molecules would not be very ordered at all. There is no strong interaction between oil molecules and water molecules, so there is no reason for water molecules to cluster around the oil molecules in any ordered way at all. If anything, pure water molecules are more likely to clustered, held together by transient dipole bonds. I'm not convinvced that DS is positive at all.

I can't say I'm an expert here, but I think you are overestimating the size of DS here. Couldn't the Boltzmanm equation give some insight here? if S=k*lnW, DS=kln(Wf/Wi) where Wf, & Wi denote the number of final and initial states possible, then even is the process increases the number of states by a huge factor like 2^1,000,000,000,000 only leads to a more modest increase in S of 1,000,000,000,000*k*ln2, and k is a pretty small constant (1.4E-23). Forgive me if I'm making a mistake here.

In my opinion there is plenty of room for DH to be large enough to outweight a small, negative DS.

Jesse, you are confusing two effects here. There are 2 entropy changes here, one caused by the physical change in the postitions of the oil and water molecules, and one caused by the change in temperature caused by this rearrangement.

Here I am saying there is an entropy increase associated with the mixing of oil and water, so the FIRST entropy change is positive. But there is a second entropy change caused by the decrease in temperature. This one is negative. (Keep in mind I am considering the process of solvation, not seperation). If the second one is larger than the first one the OVERALL entropy change is negative. The 2nd law applies to the overall change, so solvation will be non-spontaneous and the seperation will be spontaneous.

The 1st DS I am talking about is the DS in the Gibbs equation. The second DS is the DS brought about by the DH term in the gibbs equation. The total DS is DS1+DS2. DS(total) is the one which is considered in the second law, not either of the individual DS's. Note that -T*DS2=DH and DG=-T*DS(total), making this equivalent to the gibbs equation.

Jesse

Goober:
Jesse, you are confusing two effects here. There are 2 entropy changes here, one caused by the physical change in the postitions of the oil and water molecules, and one caused by the change in temperature caused by this rearrangement.

Here I am saying there is an entropy increase associated with the mixing of oil and water, so the FIRST entropy change is positive. But there is a second entropy change caused by the decrease in temperature. This one is negative. (Keep in mind I am considering the process of solvation, not seperation). If the second one is larger than the first one the OVERALL entropy change is negative. The 2nd law applies to the overall change, so solvation will be non-spontaneous and the seperation will be spontaneous.

OK, I don't think we're disagreeing here. I agree that the total entropy is a combination of the DS and the DH term--I actually said so in an earlier post. I was just confused by your statement because you just said "Seperate oil and water layers have less entropy than and oil-water mixture" without clarifying that you were just talking about the entropy associated with the positions of the molecules and not taking into account the entropy associated with the temperature change. If you are dealing with a single oil-and-water system in isolation which starts out mixed, we both agree that it will have more entropy after it becomes separated than when it was mixed, but I thought you were saying the opposite. Just a misunderstanding.

As for the issue of how important it is to consider that water molecules are more ordered when on the surface of oil, this page which I linked to earlier in the thread seems to say it's an important factor in the lower entropy of the separated state:

Supramolecular

Sorry Goober, we had a bit of a misunderstanding there re mixing or separation.

Another attractive force is the inductive force which is an induced dipole interaction between a polar molecule and a polarisable molecule (whether it be polar or non-polar). This again helps offset the slightly +ve enthalpy change due to oil and water mixing and leads to a lower overall enthalpy change when they separate.

I think you are underestimating the large +ve entropy change (and it is very positive) when oil and water separate into layers. Consider this.

Imagine two oil molecules are separated in water and lets say each one is surrounded by 8 water molecules. The reason that the water molecules around these oil molecules are 'more ordered' than when in bulk solution is because they orient themselves in such a way as to minimise and maximise the repulsive and attractive interactions between the oil and water.

Now imagine those two oil molecules come together and lower their overall surface area and are now only surrounded by 11 water molecules. This means we have the equation below:

Oil-(8H2O) + Oil-(8H2O) -----> Oil-Oil-(11H20) + H2O + H2O + H2O + H2O + H2O

i.e. 2 clusters of molecules goes to 1 cluster and 5 free water molecules, 2 going to 6. This is where the large increase in entropy comes from that drives the separation. Compared to this entropy increase, the enthalpy change can be considered close to zero and so oil-water separation is to all extent an entropy driven process. Hope this clarifies things.

Regards

Goober

OK, I see the point you're making. Do you know of any experiments anyone has done that definitely show which one is right, like any measurements of the enthapy and entropy changes?

Mine predicts that oil would be more soluble in water at higher temperatures, while yours predicts that it would become less soluble at higher temperatures. Unfortunately, the water would probably boil before you could get much difference.

I admit you are probably right, you seem to know what you're talking about more than me, but it would be nice to seem some experimental results.

Sorry, I probably should have made that clearer. My bad.

Supramolecular

Goober,

Experiments have definitely been done in calorimeters, wherein the enthalpy and free energy changes have been measured and so the entropy. I'll have a look at any references in my thesis and when I'm back in work next I can check any papers as I have full subsciptions to most journals.

Regards.

Edit.- On the topic of the solubility. As there is an endothermic component (even though it is small compared to the entropic component) oil solubility would increase with temperature.