Explaining the Immiscibility of Oil and Water: A Thermodynamic Perspective

In summary, the polarity of molecules plays a significant role in determining whether or not liquids will mix. This can be explained in terms of thermodynamics, where the free energy of mixing is entropically dominated when the A-B interaction energy is comparable to the A-A and B-B interaction energies. However, if the like interactions are stronger, there is a significant enthalpic cost to mixing, making the substances immiscible. This concept also applies to the formation of membranes at the interface. Additionally, heating a mixture of water and vegetable oil will not result in miscibility because the heat of solubility is positive and requires a negative temperature, which is impossible.
  • #1
erty
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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?
 
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  • #2
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
Danger said:
The density of two liquids determines whether or not they mix.
Then why have I been using soap all this time?
 
  • #4
Danger said:
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.
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.
 
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  • #5
Gokul43201 said:
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.

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
Gokul43201 said:
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.

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

Just asking out of curiosity..
 
  • #7
erty said:
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.
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
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
net_nubie said:
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.

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
ShawnD said:
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.


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
erty said:
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?

If the enthalpy plays no part (is 0), the only other place to get some energy is from entropy :wink:
 
  • #12
erty said:
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?
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 [itex]\Delta G = 0~so~T_c=\Delta H_{mix}/ \Delta S_{mix} [/itex]) 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 referring to a Helmholtz boundary layer?

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

Note that in this argument, we make the assumption that [itex]\Delta H [/itex] is temperature independent. This is a reasonable assumption to make in most cases, but breaks down if the temperature scale at which [itex]\Delta H [/itex] changes significantly is comparable to the critical temperature.arildno: I've got no idea what a membrane is! Are you referring to a Helmholtz boundary layer?
 
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  • #13
Gokul43201 said:
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.
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
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.
 
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  • #15
ShawnD said:
If the enthalpy plays no part (is 0), the only other place to get some energy is from entropy :wink:

But how do you know that [tex]\Delta H = 0[/tex]?
 
  • #16
Gokul43201 said:
For many liquid-liquid systems with sufficiently high boiling points, you WILL see a critical temperature (where [itex]\Delta G = 0~so~T_c=\Delta H_{mix}/ \Delta S_{mix} [/itex]) above which the two liquids are always miscible in each other and below which they are immiscible.

Yes, that makes sense.
 
  • #17
erty said:
But how do you know that [tex]\Delta H = 0[/tex]?
Are you asking about miscibility between similar liquids here?

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

Gokul43201 said:
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.
So, again, the enthalpy change during the mixing of similar liquids (i.e., polar with polar or nonpolar with nonpolar) is simplistically nothing but [itex]\Delta H(mixing) = H(mixed) - H(unmixed) = 2H(A \cdot \cdot B) - H(A \cdot \cdot A) - H(B \cdot \cdot B) [/itex].
So, if [itex]H(A \cdot \cdot B) \approx H(A \cdot \cdot A) \approx H(B \cdot \cdot B) [/itex], then [itex]\Delta H(mixing) \approx 0 [/itex].
 
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  • #18
erty said:
But how do you know that [tex]\Delta H = 0[/tex]?

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

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
Many thanks to both of you.
Now I do think I understand it, thanks for clarifying it!
 
  • #20
Gokul43201 said:
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).

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.
 
  • #21
Danger said:
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.
It only struck me later (after Russ' post) that this (mechanical mixing, rather than chemical mixing) was likely the interpretation you applied to the OP. I'm sorry my first post was curt in addressing your response to the OP.
 
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  • #22
I didn't find it all that curt; it seems like a really good post to put the facts out there and avoid confusion. :smile:
 
  • #23
ShawnD said:
Chemistry students are even told never to add water to strong acid because there's a chance it will explode.
Thought that had more to do with whether or not any liquid that splashes out will be strong acid (since adding strong acid to water is routine).
 
  • #24
cesiumfrog said:
Thought that had more to do with whether or not any liquid that splashes out will be strong acid (since adding strong acid to water is routine).

Adding 10mL of water to 1L of sulfuric acid will make so much heat that the water starts to boil. Boom. Acid all over the place, including you. It's the same idea behind adding water to boiling oil.
 
  • #25
Gokul43201 said:
It only struck me later (after Russ' post) that this (mechanical mixing, rather than chemical mixing) was likely the interpretation you applied to the OP. I'm sorry my first post was curt in addressing your response to the OP.
It isn't a big deal, I just don't like to assume a high level of understanding for someone asking a question here. I'm not sure the OP was making the distinction, so I think it was worth clarifying that the distinction existed - I mean, if s/he did understand the difference, why ask the question in the first place?
 
  • #26
russ_watters said:
I'm not sure the OP was making the distinction, so I think it was worth clarifying that the distinction existed - I mean, if s/he did understand the difference, why ask the question in the first place?
It was pretty clear from this first line in the OP, exactly what erty meant by "mixing". The entropy/enthalpy argument for solubility is an important one found in many introductory textbooks, and it was precisely this argument that the OP was unclear about. I guess I don't see what your confusion is.

Also, my previous post was directed at Danger.
 
  • #27
ShawnD said:
Adding 10mL of water to 1L of sulfuric acid will make so much heat that the water starts to boil. Boom.
I've wondered about this, and more than once, grabbed my CRC handbook and done a calculation. I've always found that adding 10mL of water to 1L of say, 10M HCl makes about as much heat as adding 10mL of 10M HCl to 1L of water. I've tried all combinations of transients from instantaneous diffusion to slow diffusion, and I've never got widely disparate results for the two cases!
 
  • #28
@ definition of mixing
I apologize for the confuzzlement. I did not know the distinction between the physical definition of mixing and the ... well, cook book's definition. This started when I mixed (oops) lemon juice and olive oil for a salad dressing, and the immiscibility thing made me wonder.
But now I know, and I will consider it the next time I talk about mixing... as for energy, work and all the other words with a very concise definition in the world of physics!

Gokul43201 said:
I've wondered about this, and more than once, grabbed my CRC handbook and done a calculation. I've always found that adding 10mL of water to 1L of say, 10M HCl makes about as much heat as adding 10mL of 10M HCl to 1L of water. I've tried all combinations of transients from instantaneous diffusion to slow diffusion, and I've never got widely disparate results for the two cases!
That's interesting. I've actually tried it, i.e. to pour water into acid. As soon as I had done it, I realized what it did, and I ran to sink while holding a bubbling beaker :)
My teacher's explanation was that the density of acid is higher than that of water, so when water is poured into a beaker of acid (I think it was 5 M HCl, so it was not that bad anyway), it does not mix well (mix?), and the acid will react violently with the water in "one layer", creating a lot of heat. If acid is poured into water, they will blend nicely before reaction, hence spreading the molecules in a even layer, delocalising the heat.
 
  • #29
erty said:
My teacher's explanation was that the density of acid is higher than that of water, so when water is poured into a beaker of acid (I think it was 5 M HCl, so it was not that bad anyway), it does not mix well (mix?), and the acid will react violently with the water in "one layer", creating a lot of heat. If acid is poured into water, they will blend nicely before reaction, hence spreading the molecules in a even layer, delocalising the heat.
That makes a lot of sense to me! Thanks; I shall rest easier with that.
 
  • #30
Gokul43201 said:
It was pretty clear from this first line in the OP, exactly what erty meant by "mixing". The entropy/enthalpy argument for solubility is an important one found in many introductory textbooks, and it was precisely this argument that the OP was unclear about. I guess I don't see what your confusion is.
Where does entropy/enthalpy come into play with an oil/water mixture? Again, my understanding here is thin, so I am in need of confirmation/correction, but it was my understanding that with a mixture of something like oil and water, there is no change in enthalpy/entropy because nothing chemical happens. It looks to me like differentiating a "mixture" from a "solution" is the entire point of the first part of the question.
 
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  • #31
russ_watters said:
Where does entropy/enthalpy come into play with an oil/water mixture? Again, my understanding here is thin, so I am in need of confirmation/correction, but it was my understanding that with a mixture of something like oil and water, there is no change in enthalpy/entropy because nothing chemical happens.
The question in the OP is: (in the above situation) why does nothing chemical happen? Or, why does oil only form a suspension (or dispersion) in water, and not a solution?

The answer follows by assuming that if it did, then there would be an increase in free energy (as explained earlier) which is unphysical. Hence, by a method of reductio ad absurdum, you prove that oil and water are immiscible.

The same argument arrives at a reduction in free energy if ethanol and water form a solution, making this the favorable final state for that system.
 
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  • #32
Gokul43201 said:
I've always found that adding 10mL of water to 1L of say, 10M HCl makes about as much heat as adding 10mL of 10M HCl to 1L of water.

That's because HCl is a gas, so "HCl solution" already has water in it.
H2SO4, the fuming kind in particular, contains absolutely no water. If you add 50mL of water to 50mL of H2SO4, the solution becomes so hot you can't even hold it with your bare hands.
This works well with sodium hydroxide too. Add water to 10M NaOH and nothing will happen. Add water to solid 99% NaOH and it will get very hot.edit:
I just realized that I didn't exactly answer your argument. You said water into acid is the same heat as acid into water, and my response was that HCl is a weak ***** acid.

To answer your question: it's different not because of the heat, but because of a difference in boiling points. Water boils at 100C, nitric acid is like 120C, and sulfuric acid is way up there past 200C (don't have an exact number). If you add sulfuric acid to water, the solution will only boil if 1). the small amount of H2S04 boils (at 200C) or 2). the entire amount of water boils (100C, but larger quantity). It takes a lot of heat to get H2SO4 up to 200C, and a lot of heat to get the majority of the water up to 100C. Adding water to sulfuric acid has the opposite rules of 1) the small amount of water boils (100C) or 2) the large amount of H2SO4 boils (200C).
Assuming both water into acid and acid into water create the same amount of heat, is it easier to reach a temperature of 100C or 200C? 100C. Adding water is bad :biggrin:
 
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  • #33
ShawnD said:
To answer your question: it's different not because of the heat, but because of a difference in boiling points. Water boils at 100C, nitric acid is like 120C, and sulfuric acid is way up there past 200C (don't have an exact number). If you add sulfuric acid to water, the solution will only boil if 1). the small amount of H2S04 boils (at 200C) or 2). the entire amount of water boils (100C, but larger quantity).
What? A solution is a single phase system. Its constituent components do not retain their pure state boiling points. When sulfuric acid dissolves in water (and unless it dissolves, there isn't any heat released), the solution acquires a new boiling point (more correctly, a tiny range of boiling points within a narrow two-phase region flanked below by the single-phase solution and above by the single-phase vapor). When you add a little sulfuric acid to a lot of water, the new BP is going to be pretty close to 100C. When you add a little water to a lot of acid, the new BP will be closer to 200C.

Either you're wrong, or I've completely misunderstood you.
 

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