# Oil, water and thermodynamics

1. Dec 23, 2006

### erty

If water and vegetable oil are added to a bowl, they do not mix.
If gasoline and vegetable oil are added in another bowl, they _do_ mix. this has something to do with the polarity of the molecules, i.e. water is polar and gasoline and vegetable oil are unpolar.

But is it possible to explain this in terms of thermodynamics?
When mixing two solutions, water and veg. oil, the entropy ought to increase, and so it should be spontaneous. Our daily life experience tells us that it is not. Why?

2. Dec 23, 2006

### Danger

Hi, erty. Welcome to PF.
Are you familiar with the concept of 'specific gravity'? The density of two liquids determines whether or not they mix. Remember that mixing is strictly a physical phenomenon, not a chemical reaction. The less dense liquid will always float on top of the other one. If they are of equal SG, they will mix.

3. Dec 24, 2006

### cesiumfrog

Then why have I been using soap all this time?

4. Dec 24, 2006

### Gokul43201

Staff Emeritus
That's not really correct. There are crude oils that are within 1% of the density of water, but won't mix with it. On the other hand, ethanol has a density that's roughly 20% lower than water's but is perfectly miscible with it. Having similar densities only helps to stabilize dispersions, not make solutions.

As erty surmised, it has a lot to do with intermolecular interactions (which are related to the polar/nonpolar nature of the substances). And miscibility can be explained in terms of thermodynamics. When the A-B interaction energy is comparable to the A-A interaction energy and the B-B interaction energy the free energy of mixing is entropically dominated (because, from the above, the enthalpy of mixing is small), making the substances miscible. When the above condition is not true (and typically, this means the like interactions are stronger than the unlike interactions), there is a significant enthalpic cost to the mixing, making the substances immiscible.

Last edited: Dec 24, 2006
5. Dec 24, 2006

### erty

Thanks.

Then if I heat the veg. oil/water-mixture, it should be possible? H = U + pV, and I assume the volumerelated work doesn't matter, so pV ~ 0, therefore H ~ U.
From G = H - TS, S > 0 when veg. oil and water are mixed together, so H < 0? That is a endothermic process, so in order to "turn it around": T(system) > T(surroundings).
We don't observe this in everyday-life, i.e. heat water and oil to a temperature that makes it miscible. So I've made an erroneous conclusion somewhere, but I can't see it myself. Help?
Or perhaps I just misunderstood the whole explanation. Either way, I don't think I understand.

6. Dec 24, 2006

### arildno

This entropic argument is directly related to whether or not there will be a membrane formation at the interface, is it not?

7. Dec 24, 2006

### net_nubie

S is change in entropy. S,therefore, is less than zero when oil and water are mixed. Also, G has to be <0 for a spontaneous process. Heat of solubility is positive for an oil-water mixture, so for the process to take place, T should be less than zero, which is impossible.

8. Dec 24, 2006

### ShawnD

Adding to goku's answer, the enthalpy is caused by differences in attraction strength between molecules. It's hard to put it into words so I'll just throw some numbers out there.

Suppose you want to mix water and ethanol. The energy needed to pull 2 water molecules apart would be maybe 100 units. The energy needed to pull 2 ethanol molecules apart would be something like 80 units. The amount of energy released by joining a water molecule to an ethanol molecule would be around 90, and this happens twice (because 2 molecules of water and 2 of ethanol). Energy required to break bonds was 100 + 80 = 180. Energy gained by forming new bonds was 90 x 2 = 180. The overall enthalpy change of mixing water and ethanol is very small, so entropy drives this reaction to happen. Water and ethanol are infinitely soluble in each other.

If you look at something like oil and water, it still takes a lot of energy to separate water molecules, but there is little to no energy gain when water joins to oil. The enthalpy is strongly against mixing of any kind.

9. Dec 24, 2006

### erty

Why is deltaS < 0 when oil and water are mixed, while deltaS > 0 when water and alcohol are mixed?
I mean, it is possible to explain this whole miscibility with polarity etc., but what is the thermodynamic explanation?

10. Dec 24, 2006

### erty

Hmm, in your example the energy required to break the intermolecular forces (I.F. to make it easier) equals the energy gained when the new I.F. are formes, so deltaS = 0, then how can the reaction be entropy-driven?

Apart from that, I understood the rest - excellent answer! Thanks.

11. Dec 24, 2006

### ShawnD

If the enthalpy plays no part (is 0), the only other place to get some energy is from entropy

12. Dec 24, 2006

### Gokul43201

Staff Emeritus
This is a fair argument, but the error is in assuming that the critical temperature where the oil-water mixture goes from being two-phase to single-phase is below the boiling point of water (assuming that boils first). We don't see the change happen becuase it takes a higher temperature for it to happen than we can make in our experiment.

For many liquid-liquid systems with sufficiently high boiling points, you WILL see a critical temperature (where $\Delta G = 0~so~T_c=\Delta H_{mix}/ \Delta S_{mix}$) above which the two liquids are always miscible in each other and below which they are immiscible.

arildno: I have no idea what a membrane is! Are you refering to a Helmholtz boundary layer?

http://www.tau.ac.il/~phchlab/experiments/Binary_Solutions/CritOpal_files/image083.jpg

Note that in this argument, we make the assumption that $\Delta H$ is temperature independent. This is a reasonable assumption to make in most cases, but breaks down if the temperature scale at which $\Delta H$ changes significantly is comparable to the critical temperature.

arildno: I've got no idea what a membrane is! Are you refering to a Helmholtz boundary layer?

Last edited: Dec 24, 2006
13. Dec 24, 2006

### Staff: Mentor

It's been a while since chemistry, but aren't you guys talking about two different things?

Mixing and dissolving are not the same - oil and water mix (not well because of specific gravity and surface tension) while water and alcohol dissolve in each other.

14. Dec 24, 2006

### Gokul43201

Staff Emeritus
I'm using the term "mixing" to mean "dissolving", as that's the usage in the OP. Strictly speaking, mixing includes both dissolving as well as dispersing (solutions and dispersions are both mixtures - the first is generally considered a homogeneous mixture, while the second is generally considered heterogeneous). But I am differentiating between solutions and dispersions (which are what you get when you have tiny droplets of A in a medium of B) though. The discussion here is about solutions, not dispersions.

Last edited: Dec 24, 2006
15. Dec 25, 2006

### erty

But how do you know that $$\Delta H = 0$$?

16. Dec 25, 2006

### erty

Yes, that makes sense.

17. Dec 25, 2006

### Gokul43201

Staff Emeritus

No, that's not correct. The energy that is used/released in breaking/making intermolecular forces is the enthalpy ($\Delta H$). The entropy is something that comes purely out of changing the "disorder" (and is not an energy). Read this bit again:

So, again, the enthalpy change during the mixing of similar liquids (i.e., polar with polar or nonpolar with nonpolar) is simplistically nothing but $\Delta H(mixing) = H(mixed) - H(unmixed) = 2H(A \cdot \cdot B) - H(A \cdot \cdot A) - H(B \cdot \cdot B)$.
So, if $H(A \cdot \cdot B) \approx H(A \cdot \cdot A) \approx H(B \cdot \cdot B)$, then $\Delta H(mixing) \approx 0$.

Last edited: Dec 25, 2006
18. Dec 25, 2006

### ShawnD

We're probably agreeing with each other, but I'm using the wrong words. This happens a lot more than I would like

Basically I'm just looking at the temperature change. If you mix water and ethanol, the temperature does not change, but the reaction still goes. Let's look at this one term at a time:

G = H - TS
G is negative (at room temp) because the reaction happens at room temperature.
H is ~0 because the temperature of the solution does not change
T is always positive; you can't have negative temperature

Then what is S?
(-) = 0 - (+)S
(-) = - (S)
Factor out the (-) and you get:
+ = S
I say this reaction is entropy driven because enthalpy does nothing. It neither helps nor hinders the reaction to any noticable extent.

I would also say that solutions with decreasing temperatures are entropy driven; this includes most kinds of salt. Ammonium Chloride is the most notable as a salt that is very soluble while the reaction is extremely endothermic.
G = H - TS
G is negative, H is large and positive, T is positive, so S must be a very large and positive.
The only reason ammonium chloride dissolves is because entropy strongly favours it. Enthalpy tries to stop this from happening.

There are also solutions which increase in temperature, such as strong acids or bases added to water. Chemistry students are even told never to add water to strong acid because there's a chance it will explode.
G = H - TS
G is negative, H is large and negative, T is positive, S might be positive but I'm not sure.
In any event, enthalpy plays a significant role in dissolving acids and bases.

19. Dec 26, 2006

### erty

Many thanks to both of you.
Now I do think I understand it, thanks for clarifying it!

20. Dec 26, 2006

### Danger

My apologies for the bad response. We were taught that 'mixing' is a strictly mechanical combination of multiple substances, with no interactions at all including disolving, polar/non-polar responses, electrostatic, hydrophobic/philic, etc., such as when you sift sand and salt together. Without regard to the substances mentioned by the OP, I was answering based upon that definition. Thanks all for putting it back on track.