Why does adding solute particles decrease the freezing point?

[Borek]

that is the brine must be infinitely diluted by condensation

Which - technically - is impossible, isn't it?

So, can the brine coexist with the ice at the water melting point?

 

[Mr Real]

Which - technically - is impossible, isn't it?

So, can the brine coexist with the ice at the water melting point?

Okay, by the equation it seems obvious that brine and ice can't coexist at the same temperature. But hadn't we earlier discussed, that condensation will occur and so the vapor pressures would become equal?

 

[Borek]

But hadn't we earlier discussed, that condensation will occur and so the vapor pressures would become equal?

Sure, trick is, water will move from ice to brine till there is no ice, and there will be only one vapor pressure - that of the brine. There will be no two different sources of the vapor.

Previous case was with two brine samples, so it was a bit different - I used that example just to make it easier for you to come to the conclusion mass can be transferred through the evaporation/condensation. Then once the concentrations became identical, process stopped. Now, the setup is different, so the final equilibrium is different as well.

 

[Mr Real]

Sure, trick is, water will move from ice to brine till there is no ice, and there will be only one vapor pressure - that of the brine. There will be no two different sources of the vapor.

Okay, so ice will disappear but I've seen that you've been emphasising the fact that it is happening at the melting point, so will ice completely disappear at melting point only, or at all temperatures? (I haven't seen the melting point play any role here so I think it will happen at every temperature, which is counterintuitive to me).

 

[Borek]

Melting point is particularly easy to understand, as it is the only point at which water and ice naturally coexist. At other temperatures predicting the outcome is a bit harder, as relative amounts of brine and ice start to play a role.

But in general we have come to the conclusion - just watching the vapor pressure is enough to explain, why brine has a lower melting point than pure water. That was the original question, wasn't it?

 

[Mr Real]

Melting point is particularly easy to understand, as it is the only point at which water and ice naturally coexist.

But didn't we just say that ice will completely disappear because of vaporization?

But in general we have come to the conclusion - just watching the vapor pressure is enough to explain, why brine has a lower melting point than pure water. That was the original question, wasn't it?

Actually, my original question was: why does freezing point decrease on adding solute to a solvent and why doesn't this decrease depend on the nature or the size of the solute particles?
Another doubt: will ice completely disappear at other temperatures also if kept in a container with brine? (Applying Raoult's law seems to suggest it will, as the formula doesn't have temperature in it)

 

[Borek]

But didn't we just say that ice will completely disappear because of vaporization?

In the presence of brine - yes. In the presence of water - no.

Actually, my original question was: why does freezing point decrease on adding solute to a solvent and why doesn't this decrease depend on the nature or the size of the solute particles?

For the same reason Raoult's law works the way it is. What matters are the molar fractions of the solute/solvent.

Another doubt: will ice completely disappear at other temperatures also if kept in a container with brine? (Applying Raoult's law seems to suggest it will, as the formula doesn't have temperature in it)

Depends on the temperature, initial brine concentration and relative amounts of ice and brine, no simple answer here.

 

[Mr Real]

Depends on the temperature, initial brine concentration and relative amounts of ice and brine, no simple answer here.

But the formula by which we got that the solution should be infinitely diluted (partial pressure formula in post #49) and as that cannot occur the ice will completely disappear; doesn't have any term for temperature, so how can it play a role here?
Now, if you've cleared the above doubt, so how would freezing point decrease, you've explained till that if we keep ice and brine in a container, the ice would disappear, so what does that have to do with the decrease in freezing point?

Thanks
Mr R

 

[Borek]

But the formula by which we got that the solution should be infinitely diluted (partial pressure formula in post #49) and as that cannot occur the ice will completely disappear; doesn't have any term for temperature, so how can it play a role here?

P₀ is a function of temperature, pressure of the saturated vapor over ice changes with the temperature as well (although at a different speed).

Now, if you've cleared the above doubt, so how would freezing point decrease, you've explained till that if we keep ice and brine in a container, the ice would disappear, so what does that have to do with the decrease in freezing point?

If there is no ice, only brine, have you reached the melting point?

What happens to the pressure of the saturated vapor over the ice when you lower the temperature? Does it go up, or down? If it goes down - what will happen when it reaches the value given by the Raoult's law for the given brine concentration (and P₀(T))?

 

[Mr Real]

My reasonings:

If there is no ice, only brine, have you reached the melting point?

If melting point is the point at which both solid and liquid states of the substance coexist, then if there is no ice that means melting point has not been reached.

What happens to the pressure of the saturated vapor over the ice when you lower the temperature? Does it go up, or down?

When the temperature is lowered, kinetic energy of the molecules decreases so vapor pressure of the ice decreases.

If it goes down - what will happen when it reaches the value given by the Raoult's law for the given brine concentration (and P0(T))?

If it reaches the value of P (which is the brine's vapor pressure), then it would coexist with the brine. But I've some doubts regarding this.
If the temperature is decreased, won't the pressure of the brine decrease too?
Also won't the Raoult's law prohibit these vapor pressures from becoming equal (like in the previous case) ?

 

[Mr Real]

Borek, can you please reply to my previous post?
Thank you
Mr R

 

[Borek]

If melting point is the point at which both solid and liquid states of the substance coexist, then if there is no ice that means melting point has not been reached.

Yes.

When the temperature is lowered, kinetic energy of the molecules decreases so vapor pressure of the ice decreases.

Yes.

If it reaches the value of P (which is the brine's vapor pressure), then it would coexist with the brine. But I've some doubts regarding this.
If the temperature is decreased, won't the pressure of the brine decrease too?

Yes, it does. It complicates things a bit (speeds at which it changes for a liquid and for a solid are different), but doesn't change the general picture.

Also won't the Raoult's law prohibit these vapor pressures from becoming equal (like in the previous case) ?

Quite the opposite - it will allow you to find such an x that these pressures are identical for a given temperature. This will give the composition of the brine that has the meting point at the given temperature.

Borek, can you please reply to my previous post?

Can I get back home?

 

[Mr Real]

Yes, it does. It complicates things a bit (speeds at which it changes for a liquid and for a solid are different), but doesn't change the general picture.

So, the decrease in vapor pressure must be less for the brine than for the ice(because only then the vapor pressure of the ice can equal the pressure over the brine), isn't it? If so, What is the reason behind it?

Quite the opposite - it will allow you to find such an x that these pressures are identical for a given temperature. This will give the composition of the brine that has the melting point at the given temperature.

So, both the vapor pressures become equal and the ice and the brine coexist. Why do we need to find x, then?

Can I get back home?

Sorry, I thought I had become too irksome (So you decided not to reply). : )

 

[Borek]

So, the decrease in vapor pressure must be less for the brine than for the ice(because only then the vapor pressure of the ice can equal the pressure over the brine), isn't it? If so, What is the reason behind it?

Actually I think it is the other way - vapor pressure over the liquid goes down faster. But I am not sure. It is all about the interactions between molecules in the liquid and in the solid.

So, both the vapor pressures become equal and the ice and the brine coexist. Why do we need to find x, then?

Because what we are interested in is the dependence between the melting point and the salt concentration (given by x). We can either start with a temperature, check vapor pressures and calculate x, or use x to calculate vapor pressure over the solution and then check at what temperature this vapor pressure is identical to the vapor pressure over solid. In both cases we get a pair of (x,T).

 

[Mr Real]

Because what we are interested in is the dependence between the melting point and the salt concentration (given by x). We can either start with a temperature, check vapor pressures and calculate x, or use x to calculate vapor pressure over the solution and then check at what temperature this vapor pressure is identical to the vapor pressure over solid. In both cases we get a pair of (x,T).

So, I got what you said and also that you were using this system, consisting of ice and brine as a simple analogy, so now how does it all fit together when I add salt to a solvent and its freezing point decreases, because here we don't have ice and brine separately, do we? And I don't think that at freezing point some part of it behaves as brine and the other as ice or does it?
Another question: you had said that the size of the solute particles doesn't matter in the depression of freezing point for the same reason it doesn't matter in Raoult's law, so why doesn't it matter in Raoult's law?

 

[Borek]

So, I got what you said and also that you were using this system, consisting of ice and brine as a simple analogy, so now how does it all fit together when I add salt to a solvent and its freezing point decreases, because here we don't have ice and brine separately, do we? And I don't think that at freezing point some part of it behaves as brine and the other as ice or does it?

Come one, this one is trivial. Can you have the ice in the presence of brine if the temperature is not low enough? If you can't - doesn't it mean the melting point is somewhere at the lower temperature?

Another question: you had said that the size of the solute particles doesn't matter in the depression of freezing point for the same reason it doesn't matter in Raoult's law, so why doesn't it matter in Raoult's law?

Good question, one that I don't know exact answer to. But remember that science rarely (or never) answers questions about "why". It answers questions about "how".

 

[Mr Real]

Come on, this one is trivial. Can you have the ice in the presence of brine if the temperature is not low enough? If you can't - doesn't it mean the melting point is somewhere at the lower temperature?

I know this but that's not what I meant. I meant that can you explain why freezing point decreases in reference to the original question I had asked, that of why does adding solute to a solvent decrease its freezing point? I got the reason for the hypothetical system which consisted of ice and brine kept separately, but I want to know how to explain this in the actual context where we are adding a solute, say salt to a solvent, say water and observing at what temperature this solution freezes, here we don't have that hypothetical system containing ice and brine kept separately.

remember that science rarely (or never) answers questions about "why". It answers questions about

Okay, then how is it that the nature (especially size) of solute particles doesn't play any role in the Raoult's Law. ; )
Well, is it because pressure doesn't depend on size only number of molecules? (e.g. the pressure exerted by 50 H2 molecules is the same as that exerted by 50 CO2 molecules, even though H2 is very small in size comparatively). If it is the reason, then do you know why size of molecules doesn't matter for pressure?
( Also is pressure more closely related to mvelocity, kinetic energy or momentum of the molecules?)

Thank you so much
Mr R

 

[Borek]

I know this but that's not what I meant. I meant that can you explain why freezing point decreases in reference to the original question I had asked, that of why does adding solute to a solvent decrease its freezing point? I got the reason for the hypothetical system which consisted of ice and brine kept separately, but I want to know how to explain this in the actual context where we are adding a solute, say salt to a solvent, say water and observing at what temperature this solution freezes, here we don't have that hypothetical system containing ice and brine kept separately.

But there is no need for other explanation. You have shown (using the hypothetical case) why ice can't exist in the presence of brine. But if it can't exist in the hypothetical case when we start with a brine and a piece of ice, it won't exist in other cases, when all we have is a brine at the given temperature. It will appear when the temperature goes down, which is what the melting point depression is about.

Well, is it because pressure doesn't depend on size only number of molecules? (e.g. the pressure exerted by 50 H2 molecules is the same as that exerted by 50 CO2 molecules, even though H2 is very small in size comparatively). If it is the reason, then do you know why size of molecules doesn't matter for pressure?
( Also is pressure more closely related to mvelocity, kinetic energy or momentum of the molecules?)

This is getting speculative. Besides so far we were talking about ideal solutions - when it comes to the real ones things get wacky quite fast, and you better leave that now, as you have enough conceptual problems with the idealized case.

Sorry to be a show stopper. To be honest I am not convinced I know thermodynamics of the real solutions well enough to be ready for a detailed discussion. Besides, these things get quite math heavy fast.

 

[Mr Real]

It will appear when the temperature goes down, which is what the melting point depression is about.

Yes, that's precisely what I'm saying, that at melting point, ice does appear. So, in this real case, I don't think that some part of solvent starts behaving as ice and the other as brine, so what happens in this real scenario. How can all the reasoning we have applied for the hypothetical case be applied here?
Because if I'm required to explain this occurrence(ex: to a teacher or in an exam), then I would naturally be expected to give/write an explanation for the real case.

 

[Borek]

If you start with a brine (which have some vapor pressure, given by the Raoult's law) and you start lowering temperature, at what point will the ice appear? (think in terms of vapor pressures, as we did in all earlier posts).

 

[Mr Real]

If you start with a brine (which have some vapor pressure, given by the Raoult's law) and you start lowering temperature, at what point will the ice appear? (think in terms of vapor pressures, as we did in all earlier posts).

I think as we lower the temperature, the vapor pressure of the brine will go on decreasing, and when just when it becomes equal to the vapor pressure of ice, the ice will appear and this temperature will be the melting point.

 

[Borek]

I think as we lower the temperature, the vapor pressure of the brine will go on decreasing, and when just when it becomes equal to the vapor pressure of ice, the ice will appear and this temperature will be the melting point.

While what you wrote is generally true, it is also very subtly wrong, as if you didn't got what the whole discussion was about The only stable phase is the one with the lower vapor pressure. Ice doesn't appear when the vapor pressure over brine will decrease to the one over ice, ice could nicely exist when the pressure over brine was higher. Actually if the pressure over the brine was higher than the one over ice water would nicely evaporate from the brine and solidify as ice.

In other words: it is not decrease of the saturated vapor pressure over the brine that is the driving force here, it is the decrease of the saturated vapor pressure over the ice that allows ice to appear at some point.

 

[Mr Real]

While what you wrote is generally true, it is also very subtly wrong, as if you didn't got what the whole discussion was about The only stable phase is the one with the lower vapor pressure
Ice doesn't appear when the vapor pressure over brine will decrease to the one over ice, ice could nicely exist when the pressure over brine was higher. Actually if the pressure over the brine was higher than the one over ice water would nicely evaporate from the brine and solidify as ice.

In the context of the hypothetical system we were earlier considering, I got that the vapor pressure over ice must decrease so that it can become equal to the vapor pressure of the brine. But in this case, we only have brine to start with, no ice is present initially; so as we decrease the temperature, doesn't the vapor pressure of the brine go on decreasing until at the melting point, the vapor pressure of the brine becomes equal to that of the ice?

In other words: it is not decrease of the saturated vapor pressure over the brine that is the driving force here, it is the decrease of the saturated vapor pressure over the ice that allows ice to appear at

But if ice isn't even present in the beginning in this case, then how can it have a vapor pressure and so how can this non-existent vapor pressure decrease?

 

[Borek]

It doesn't matter that there is initially no ice present. The moment the vapor pressure will meet that of the ice (at the given temperature), it will condense producing ice.

 

[Mr Real]

It doesn't matter that there is initially no ice present. The moment the vapor pressure will meet that of the ice (at the given temperature), it will condense producing ice.

Okay, the vapor pressure of the brine becomes equal to the vapor pressure that ice has at that temperature but in this process at what step did/will the saturated vapor pressure of the ice decrease?

 

[Borek]

Why should it matter? Saturated vapor pressure over the ice at the melting point is around 0.61 kPa, it is a property, you don't need the presence of the ice for this number to exist.

 

[Mr Real]

Why should it matter? Saturated vapor pressure over the ice at the melting point is around 0.61 kPa, it is a property, you don't need the presence of the ice for this number to exist.

Okay, so in the whole process of vapor pressure of the brine decreasing with decrease in temperature, so as to become equal to that of the ice at melting point, at what step did/will the saturated vapor pressure of the ice decrease?

 

[Borek]

Sorry, I have no idea what you are asking about. Apparently you are confused about something, but I have no idea about what.

 

[Mr Real]

Sorry, I have no idea what you are asking about. Apparently you are confused about something, but I have no idea about what.

I'm confused about what you said in post #72 which is:

it is not decrease of the saturated vapor pressure over the brine that is the driving force here, it is the decrease of the saturated vapor pressure over the ice that allows ice to appear at some point

So, this is my doubt, how can the decrease in the vapor pressure of the ice the driving factor in this scenario, where we get a salt solution and start decreasing the temperature, and as the temperature decreases, the brine's vapor pressure decreases and at the melting point it becomes equal to the vapor pressure of ice. Where did ice's vapor pressure decrease in all this process? Isn't the decrease in brine's vapor pressure the driving force?

 

[Borek]

If you assume, to simplify things, that P₀ (saturated vapor pressure over water) is constant and is not a function of the temperature, you get this:

We start right to the melting point (only brine present) and we lower the temperature. When temperature goes down, the pressure of the saturated vapor pressure over ice goes down (there is no ice present, but it doesn't matter - we know what value the saturated pressure would have, that's what the curve tells us). When it gets to the point where the pressure over ice curve meets the pressure over brine, the ice will appear.

In reality P₀ is a function of the temperature as well, but it has a different curvature, so two curves meet at some point.

 

[Mr Real]

In reality P₀ is a function of the temperature as well, but it has a different curvature, so two curves meet at some point.

So, doesn't the appearance of ice depend on decrease of vapor pressures of ice and brine both?
Another question: In the diagram, is the point marked as "melting point" ,the decreased melting point or the original one, i.e. 0° celsius?

 

[Borek]

So, doesn't the appearance of ice depend on decrease of vapor pressures of ice and brine both?

In a way, but it is still wrong approach. It is not "decrease of the vapor pressures" that is what is important here, but the fact that they get equal. Yes, they get there by going down, but in other cases vapor pressures go up and meet at the phase change.

Another question: In the diagram, is the point marked as "melting point" ,the decreased melting point or the original one, i.e. 0° celsius?

Lowered one, see the pressure axis - where is P₀? where (and what) is P₀(1-x)?

 

[Mr Real]

In a way, but it is still wrong approach. It is not "decrease of the vapor pressures" that is what is important here, but the fact that they get equal. Yes, they get there by going down, but in other cases vapor pressures go up and meet at the phase change.

In post #72, you said that the driving force behind the appearance of ice is the decrease in vapor pressure of ice, that's what I'm asking: isn't the decrease in vapor pressures of both ice and brine responsible for the appearance of ice, rather than just that of the ice?

Lowered one, see the pressure axis - where is P₀? where (and what) is P₀(1-x)?

Yeah, it is the reduced melting point. (P₀ is above P₀(1-x). P₀(1-x) is the vapor pressure of the brine at the reduced melting point). I have a question, will the curve of v.p. over the brine be steeper than the curve of ice above the melting point, or not?
Another doubt, shouldn't there be a break in the curve representing v.p. over brine (because after melting point is reached, if we continue on decreasing the temperature, then the liquid state (i.e. the brine) will no longer exist as only the solid state exists below the melting point) am I correct?

 

[Borek]

In post #72, you said that the driving force behind the appearance of ice is the decrease in vapor pressure of ice, that's what I'm asking: isn't the decrease in vapor pressures of both ice and brine responsible for the appearance of ice, rather than just that of the ice?

OK, I see what you mean. On some level you are perfectly right - yes, the ice appears when its vapor pressure gets sufficiently low, and yes, the saturated vapor pressure over the brine goes down as well. Problem I have with your wording is that I have a feeling (perhaps a wrong one) you are concentrating on a single fact ("pressure of ice vapor decreases") ignoring whole process involved (evolution of the system in the direction of a possible equilibrium between phases, no matter what the starting point is).

I have a question, will the curve of v.p. over the brine be steeper than the curve of ice above the melting point, or not?

The higher the ΔH_vap, the stepper the curve. Simple application of the Clausius-Clapeyron equation.

Another doubt, shouldn't there be a break in the curve representing v.p. over brine (because after melting point is reached, if we continue on decreasing the temperature, then the liquid state (i.e. the brine) will no longer exist as only the solid state exists below the melting point) am I correct?

Yes and no. Yes - your logic about the brine disappearance is correct. But, first, we can estimate what the saturated vapor pressure would be (using the Clausius-Clapeyron equation and assuming ΔH_vap doesn't depend on the temperature - not a bad assumption for small temperature changes), second, if we manage to supercool the brine liquids we can just measure the pressure experimentally. Tricky and difficult, but doable.

 

[snorkack]

And "brine" is something that actually does not have a constant melting point because it does not have a constant composition.
Consider the following examples:

1. In a sealed vessel, there are two open vessels: one of water, the other of brine. Then as long as water is present, its vapour pressure will always be lower than that of brine. Meaning that water vapour will evaporate from water vessel and condense into brine vessel, until the water vessel is completely dry. Only the speed of the process and the quantity left as vapour in the airspace of the sealed vessel will depend on temperature.
2. In a sealed vessel, there are two open vessels: one of more concentrated brine, the other of more dilute brine. Now, the vapour pressure of each vessel will still differ at any temperature, but it does not depend on temperature alone - also of concentration. The water vapour will evaporate from the vessel of diluter brine and condense into vessel of more concentrated brine, but since there is salt in the vessel that originally held diluter brine, it cannot evaporate to dryness - the process goes on until the concentrations of brine in both vessels are equal.
3. In a sealed vessel, there is an open vessel of water, and also a crystal of ice. The vessel is below freezing point of water, but the crystal of ice may be out of contact with water. What will happen? Well, the liquid water would be supercooled and therefore its vapour would be supersaturated. Water would evaporate from supercooled liquid, and condense to the hoarfrost crystal out of contact with water, until the water vessel is dry as in case 1), and all water has turned into ice. The outcome - all water turns into ice - is the same whether the ice is out of contact with water and grows by condensation, or whether the water is in contact with ice and undergoing freezing directly. All that is necessary is the temperature (water is below freezing point) and that there is some ice available to condense. Only in absence of ice nuclei would both supercooled water and supersaturated vapour be stable.
4. In a sealed vessel, there is an open vessel of dilute brine, and also crystals of ice. Then, depending on the exact temperature and concentration of the brine, there are two possible outcomes: either the vapour pressure of brine is lower, in which case ice sublimes and condenses into brine, diluting the brine and increasing its vapour pressure... until brine comes to an equilibrium at a certain concentration. Or else the vapour pressure of ice is lower, in which case brine evaporates and condenses to ice, concentrating the brine and lowering its vapour pressure... until, again, equilibrium is reached at a certain brine concentration.

 

[Borek]

All that is necessary is the temperature (water is below freezing point) and that there is some ice available to condense.

Oh my. We have discussed similar examples to those you have listed in the thread earlier (have you read them?), now you have added another level of complication to the system.

 

[Mr Real]

The higher the ΔH_vap, the stepper the curve. Simple application of the Clausius-Clapeyron equation.

So, was I correct when I said that the vapor pressure of the brine will steeper than that of the ice?
If I was correct,then I think I have got it, let's see if I can sum it up: Normally, if we take water it freezes at 0° C, because the vapor pressures of ice and water become equal at this temperature. But when we add salt, its vapor pressure decreases (We can see it by Raoult's law), and if now we take it to 0° C, it doesn't freeze(again we can see it from the Raoult's law). So now what we'll have to do is, we'll have to decrease the temperature because of which the saturated vapor pressures over the brine as well as the ice would decrease, but (I think this fact is of Prime Importance), the decrease over brine would be less than the decrease over ice (because only then the two vapour pressures can become equal), so ice would appear at this decreased melting point, so freezing point has been reached. Was I wrong anywhere?
Thanks
Mr R

 

[Borek]

Looks good (unless I am missing something).

This is calculated from Claussius-Clapeyron equation assuming P₀=611 Pa (not exactly what we need, this is a triple point, but close enough), ΔH_vap=45.05 kJ/mol, ΔH_sub=51.06 kJ/mol, 1 molal glucose solution (better than brine, no need for Van 't Hoff factor) which has x=0.0177 (after taking into account solution density).

 

[Mr Real]

Looks good (unless I am missing something).

View attachment 204806

This is calculated from Claussius-Clapeyron equation assuming P₀=611 Pa (not exactly what we need, this is a triple point, but close enough), ΔH_vap=45.05 kJ/mol, ΔH_sub=51.06 kJ/mol, 1 molal glucose solution (better than brine, no need for Van 't Hoff factor) which has x=0.0177 (after taking into account solution density).

*I'll pretend that the we're still talking about brine rather than glucose, as I feel more comfortable about it*
From this graph, I see that the curve for the ice is steeper than the curve for brine, that means before the melting point is reached, the decrease in vapor pressure is more for the brine than for the ice(so I was wrong), then how can the two vapour pressures even become equal? I can't make heads or tails of it. Hope you see my confusion.

better than brine, no need for Van 't Hoff factor

Out of curiosity, how will the Van't Hoff factor affect all this?

Thanks

 

[snorkack]

Oh my. We have discussed similar examples to those you have listed in the thread earlier (have you read them?), now you have added another level of complication to the system.

Yes, I read them. It seemed to me that several important points were still being missed, to judge from the responses.

This is calculated from Claussius-Clapeyron equation assuming P₀=611 Pa (not exactly what we need, this is a triple point, but close enough), ΔH_vap=45.05 kJ/mol, ΔH_sub=51.06 kJ/mol, 1 molal glucose solution (better than brine, no need for Van 't Hoff factor) which has x=0.0177 (after taking into account solution density).

A confusing factor here is the failure to terminate the ice vapour pressure curve at 273,16 K.
Also, it would be illustrative to include the vapour pressure curve of pure (supercooled) water.

As for Van´t Hoff factor: since brine undergoes electrolytic dissociation into Na⁺ and Cl^- ions, its concentration is about twice bigger than the amount of NaCl units... but, since some of the ions form ion pairs, not quite twice. The actual Van´t Hoff factor varies. Something that does not happen with glucose, which neither dissociates nor associates.

 

[Borek]

*I'll pretend that the we're still talking about brine rather than glucose, as I feel more comfortable about it*
From this graph, I see that the curve for the ice is steeper than the curve for brine, that means before the melting point is reached, the decrease in vapor pressure is more for the brine than for the ice(so I was wrong), then how can the two vapour pressures even become equal? I can't make heads or tails of it. Hope you see my confusion.

I prepared the plot just so that you can follow the pressures and analyze them. Have you tried?

Out of curiosity, how will the Van't Hoff factor affect all this?

It will just muddy the water (no pun intended) adding another variable. For glucose solution it simply equals 1.

 

[Borek]

A confusing factor here is the failure to terminate the ice vapour pressure curve at 273,16 K.
Also, it would be illustrative to include the vapour pressure curve of pure (supercooled) water.

Feel free to make a better one.

 

[Mr Real]

I prepared the plot just so that you can follow the pressures and analyze them. Have you tried?

Oh, analysing the graph more carefully, I see that the decrease in vapor pressure is less for glucose solution than for the solid state for the same decrease in temperature, so I was right earlier when I said that the decrease in vapor pressure will be less for the brine than that for the ice. And, will this be true for all solvents?

Thanks so much
Mr R

 

[Borek]

will this be true for all solvents?

As I signaled earlier - how steep the curve is depends on the ΔH_{getting into the gas phase}. Note that ΔH_vap and ΔH_sub describe very similar process - you have a condensed phase that turns into a gas. To do so, molecules of the condensed phase have to break all bonds with molecules they are surrounded by, so it turns out ΔH_{getting into the gas phase} is a function of the strength of the interactions between molecules. These interactions are always stronger in solids than in liquids.

 

[snorkack]

To do so, molecules of the condensed phase have to break all bonds with molecules they are surrounded by, so it turns out ΔH_{getting into the gas phase} is a function of the strength of the interactions between molecules. These interactions are always stronger in solids than in liquids.

With an extremely rare exception, that is generally a very poor solvent as well. He has negative enthalpy of fusion... pure He-4 under 0,77 K and pure He-3 under 0,3 K.
Are He-4 and He-3 miscible in solid phase? How is freezing point of He-4 affected by dissolved He-3?

 

[Mr Real]

As I signaled earlier - how steep the curve is depends on the ΔH_{getting into the gas phase}. Note that ΔH_vap and ΔH_sub describe very similar process - you have a condensed phase that turns into a gas. To do so, molecules of the condensed phase have to break all bonds with molecules they are surrounded by, so it turns out ΔH_{getting into the gas phase} is a function of the strength of the interactions between molecules. These interactions are always stronger in solids than in liquids.

So, is it true for all solvents? I think it is because ΔH_sub is always greater than ΔH_vap(or is it not?). So, the sublimation curve will always be steeper than the vaporization curve.

What I'm asking is this: I don't think the vapor pressures can become equal if the decrease in vapor pressure is more for the solution than for the solid state of the solvent, can it?

 

[Borek]

I don't think the vapor pressures can become equal if the decrease in vapor pressure is more for the solution than for the solid state of the solvent, can it?

Instead of discussing with facts follow the plot starting from 0°C and see what is going on.

 

[Mr Real]

Instead of discussing with facts follow the plot starting from 0°C and see what is going on.

Okay, I've written all my conclusions:

• From the graph, I see that after 0°C, the decrease in v.p. of the 1M glucose solution curve is lower than that for the sublimation curve.
• Also the curve for sublimation is steeper than the curve for glucose.
• I also get the following conclusions(I don't think they are of importance for our problem); that because the curve for sublimation has lower v.p. so, the solid state will be the stable one below the freezing point(obviously). Above the freezing point, the curve for 1M glucose solution has less v.p. so it is the stable state (again obvious).

 

[Borek]

From the graph, I see that after 0° C, the decrease in v.p. of the 1M glucose solution curve is lower than that for the sublimation curve. Also the curve for sublimation is steeper than the curve for glucose.

While these are true you have ignored the most important thing the plot shows.

Move from 0°C to lower temperatures - does the difference between pressures go down, or up?

 

[Mr Real]

.
Move from 0°C to lower temperatures - does the difference between pressures go down, or up?

The difference between the pressures goes up as we move from 0°C to lower/upper temperatures.