Why does adding solute particles decrease the freezing point?

[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.

 

[Borek]

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

No, you are misreading the plot.

Temperature scale is in K, so the zero is at 273.16.

Besides, if what you wrote were true, it would be impossible to get to the point where both pressures are identical.

I have marked for you pressure differences at 266K and 268K - which one is greater?

 

[Mr Real]

No, you are misreading the plot.
Temperature scale is in K, so the zero is at 273.16.

No, I'm not misreading the plot and I know that the temperature scale is in K.(Besides, there is no point marked as 0 in the graph, so how can I get confused about it )

Besides, if what you wrote were true, it would be impossible to get to the point where both pressures are identical.

Well, I think what I wrote is true. Maybe I didn't put clearly what I mean there. I mean that from the graph we can see that if we go to lower temperatures from 0°C (i.e. 273.16 K), the difference in v.p. becomes larger and larger. Similarly, from the graph I can also see that if we go to upper temperatures from 0°C, the difference in v.p. becomes larger and larger (e.g. the difference in v.p. is larger for 276 K than for 274 K), isn't that right?

I have marked for you pressure differences at 266K and 268K - which one is greater?

The pressure difference at 266K is greater.

 

[Borek]

I mean that from the graph we can see that if we go to lower temperatures from 0°C (i.e. 273.16 K), the difference in v.p. becomes larger and larger.

Quite the opposite, at least initially.

 

[Mr Real]

Quite the opposite, at least initially.

View attachment 204866

Yes, I can see that but What I meant is this: If we start from 273K and go to lower temperatures like 272 K, 270 K, etc. Tthe difference in vapor pressure is becoming larger and larger. And similarly if we start from 273 K and go up to higher temperatures, like 275 K, 276 K etc. the difference in vapor pressure is again getting larger and larger. The point is I meant we start from 273 K. (See I always said we go from 0°C to lower/higher temperatures)

 

[Borek]

Sorry, I have no idea what you mean.

First, the difference (understood as V_vap-V_sub) is getting lower and lower when we move to higher temperatures, see the first plot.

Second, what really matters for us here, is the absolute value of the difference (second plot), as as long as the pressures are not equal system is not at the equilibrium, and it is the equilibrium that we are interested in (as this is where the new melting point is). Absolute value for the 1 molal solution for which the calculations were done has minimum at around 271.3 K (normal melting point minus cryoscopic constant). (Minimum value should lie on zero, just the plot is inaccurate). So, if we move from 273 down, the difference gets lower till it hits zero, it doesn't grow, at least initially.