[iScience]

in my notes it says the following about freezing point depression.

As a solvent freezes, molecules rearrange themselves into the positions to which the crystalline is formed. When foreign constituents of the solution (ie particles other than solvents, ie solutes) is introduced to a freezing solvent, the solvent particles now have to form their structures around the foreign solute particles which requires a lower thermal energy state than would the pure solvent.

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However... if my solvent constituents were relatively non-polar and my solute constituents were relatively polar enough to a certain degree, wouldn't the freezing point increase as the impure crystal would no longer require the temperature to be low enough since the polarity takes care of the rest?

[gritch]

Freezing point depression is fundamentally due to entropic effects. By adding a solute you increase the entropy of the overall system, causing a greater enthalpic barrier (ie you have to cool a system more) to create a more ordered system. In an ideal system, one in which there is no intermolecular interactions in the solution at all, one would still expect a freezing point depression on an entropic basis alone.

I threw out a lot of thermodynamic terms there but basically it means it takes more energy to organize a more disordered system than an ordered one.

However... if my solvent constituents were relatively non-polar and my solute constituents were relatively polar enough to a certain degree, wouldn't the freezing point increase as the impure crystal would no longer require the temperature to be low enough since the polarity takes care of the rest?

I don't exactly see what you're trying to say here. The amount of solute present should be vastly lower than the amount of solvent. By the point you'd have enough solute molecules to increase the freezing point it wouldn't really be considered a solute at that point. I doubt the effect of polarity is enough to offset the entropic increase at concentrations where calling the substance a solute is appropriate.

We do need to make sure however that while freezing we actually are observing true solution behavior. If solute and solvent become in-miscible at a certain temperature we need to consider the the solution as a two phase system instead now. It's also possible for complexes between solute and solvent to form at different temperatures (maybe like hydration shells around a metal ion) that might throw off our observations as well.

[curiouscat]

Freezing point depression is fundamentally due to entropic effects. By adding a solute you increase the entropy of the overall system, causing a greater enthalpic barrier (ie you have to cool a system more) to create a more ordered system.

Out of curiosity, what about a BP elevation?  Is that also entropic?

[gritch]

Out of curiosity, what about a BP elevation?  Is that also entropic?

Yes, I believe boiling point elevation is also based on entropic considerations as well. I'm finding it a bit tougher to explain than freezing point so bear with me here.

Let's look at the concept of boiling itself - or rather just going from the liquid to gas phase. There is a favorable entropic drive to go from the liquid to gas phase (after all gases are generally more disordered than liquids) but in order to transfer between these phases we must overcome the intermolecular forces between molecules in the liquid phase - the stronger these forces the higher the boiling point and so on.

By adding a solute we don't increase the intermolecular forces within the solution to any appreciate amount (remember we have very little solute) but we do increase the entropy of the solution. Now the difference between the entropy of being in the gas phase and the entropy of the solution has been lowered. We still have to overcome the same intermolecular forces the as before to achieve boiling but now it's less favorable to go into the gas phase in the first place, thus we have to put more energy in to maintain the gas phase. This is what leads to a boiling point elevation.

[iScience]

sadly my question has not yet been answered      (at least not to my satisfaction)

let's speak entirely outside the domain of thermodynamics; i explained the direct systematic process of WHY the freezing point depresses with the addition of solutes. i don't count entropy explanations as a direct explanation for why these things happen. As i've stated, the freezing point depresses because as a liquid freezes and forms a crystalline structure, when a solute is introduced into the solvent, the crystallizing solvent must now crystallize around the solvent which is energetically less favorable in terms of bond formation due to the now uneven geometries of the less morphous crystalline structure and therefore requires the temperature to be lower to actually form a crystal around the solute particles.

restating my question, if the solutes were sufficiently polar, would it have a counter effect on the default effects of solutes in a liquid? (ie the freezing point depression)

[curiouscat]

let's speak entirely outside the domain of thermodynamics; i explained the direct systematic process of WHY the freezing point depresses with the addition of solutes. i don't count entropy explanations as a direct explanation for why these things happen.

Then you are mistaken. That's like asking "Why does an apple fall from a tree? But please no discussions that involve Newton's Laws or gravity!"

restating my question, if the solutes were sufficiently polar, would it have a counter effect on the default effects of solutes in a liquid? (ie the freezing point depression)

Why don't you search for a concrete counter example then? Do you know of any solute examples that illustrate your hypothetical? I don't (but I may be wrong).

[iScience]

let's speak entirely outside the domain of thermodynamics; i explained the direct systematic process of WHY the freezing point depresses with the addition of solutes. i don't count entropy explanations as a direct explanation for why these things happen.

Then you are mistaken. That's like asking "Why does an apple fall from a tree? But please no discussions that involve Newton's Laws or gravity!"

restating my question, if the solutes were sufficiently polar, would it have a counter effect on the default effects of solutes in a liquid? (ie the freezing point depression)

Why don't you search for a concrete counter example then? Do you know of any solute examples that illustrate your hypothetical? I don't (but I may be wrong).

with the falling apple, i can ask well why does gravity have an effect on the mass of the apple? i see no apparent reason, and when i go back far enough i can go back to an explanation provided by the Higgs field.
For an electric potential field, i can say that well, this electrostatic potential exists because of the fact that charged particles are influenced by electric fields and due to this fact, the  -dir(gradient of the electric field) is the field of electrostatic potential.
but with entropy.... the number of ways you can rearrange a system (W). this does not tell me anything directly. I could ask how a sodium-potassium pump works, and one could say.. because it follows... the entropy trend or something of the sort, or simply, "because of entropy", and this would tell me exactly nothing at all. biology students learn the PROCESS behind how it works, what happens step by step mechanically, and not explanations of "entropy increases or decreases".
please do not mistake what i am saying for an insult to the subject of thermodynamics. i'm sure the concept of entropy has its place in certain subjects. However the replies so far do not address my question. something being "due to entropy and therefore causing a higher enthalpic barrier" only gives me a mathematical model/framework of the system and i have no way of relating this mathematical framework back to what is visually actually going on.

for the concrete example, if there was one i wouldn't be compelled to ask for the possibility. i enjoy working in the theoretical domain, asking what ifs, so that's what i was doing rather.