Why does adding solute particles decrease the freezing point?

[Mr Real]

OK, so now try to predict what will happen if you put a piece of ice and a beaker of brine, both at 0°C, in the same container.

I already told you to consider vapor pressures of both, please don't force me to move your legs one by one at each step.

Then, we would obtain the same vapor pressure over both of them.

 

[Borek]

Then, we would obtain the same vapor pressure over both of them.

Are the initial pressures identical?

 

[Mr Real]

Are the initial pressures identical?

No, the initial pressure is less over the ice compared to the brine (because ice is a solid whereas brine is a solid-liquid solution).

 

[Borek]

No, the initial pressure is less over the ice compared to the brine (because ice is a solid whereas brine is a solid-liquid solution).

Think it over.

At 0°C vapor pressure above ice and pure water are identical, which makes it possible for water and ice to coexist at the melting point.

What happens with vapor pressure over the water, when you add salt? Apply the Raoult's law.

 

[Mr Real]

What happens with vapor pressure over the water, when you add salt?

Oh, the vapor pressure of water will decrease on adding salt. So, the initial vapor pressure of ice will be greater than that of brine (seems very odd to me).

Apply the Raoult's law.

But I have read Raoult's law is applicable only for liquid-liquid solutions and this (brine) is a solid-liquid solution we are talking about, isn't that correct?

 

[Borek]

Oh, the vapor pressure of water will decrease on adding salt. So, the initial vapor pressure of ice will be greater than that of brine.

Yes. Now, remember that part about mass transfer when the equilibrium vapor pressures are different? Will the ice survive in such conditions, or will it evaporate and condense diluting the solution? In other words: is it possible for the ice to coexist with a brine?

But I have read Raoult's law is applicable only for liquid-liquid solutions and this (brine) is a solid-liquid solution we are talking about, isn't that correct?

Raoult's law works works for every mixture (well, for every ideal mixture, but it doesn't matter much here). However, as the vapor pressure over the solid is typically negligible, for liquid-solid mixtures we use it to calculate vapor pressure of the solvent only.

 

[Mr Real]

Yes. Now, remember that part about mass transfer when the equilibrium vapor pressures are different?Will the ice survive in such conditions, or will it evaporate and condense diluting the solution?

So, as the initial vapor pressure over the brine is lower than that over the ice therefore condensation will occur over the brine equalizing both vapor pressures and diluting the brine. So I know the ice will vaporize but I don't know whether it will survive because I don't know just what amount of the ice will evaporate(as the ice will disappear only when it evaporates completely and I don't know why that will happen).

In other words: is it possible for the ice to coexist with a brine?

According to me, it is possible.

the vapor pressure over the solid is typically negligible

Yeah, seems pretty weird to me that why did ice have more initial vapor pressure than the brine (I've got the reasoning but as I have said it seems very odd to me that a solid has greater v.p. than a liquid)

 

[Borek]

So, as the initial vapor pressure over the brine is lower than that over the ice therefore condensation will occur over the brine equalizing both vapor pressures and diluting the brine.

Right. That means ice disappears in the presence of brine, so as long as the brine is present there is no ice at the melting point, and to solidify brine we have to lower the temperature till the vapor pressure over the brine is equal to that over the ice, yes?

So I know the ice will vaporize but I don't know whether it will survive because I don't know just what amount of the ice will evaporate(as the ice will disappear only when it evaporates completely and I don't know why that will happen).

No matter how much ice evaporates, is - ever - the brine diluted infinitely, or is - always - salt concentration finite? And as long as the concentration is finite, is the vapor pressure over the brine equal, or lower to the vapor pressure over the ice (we are talking Raoult's law all the time here)?

Yeah, seems pretty weird to me that why did ice have more initial vapor pressure than the brine (I've got the reasoning but as I have said it seems very odd to me that a solid has greater v.p. than a liquid)

We are talking about two different solids here. Ice is a solid made of small, covalent molecules, which makes it quite volatile. Ionic salts (like NaCl) which are what is dissolved in a brine, have the vapor pressure many orders of magnitude lower. I guess you were told about latter being non-volatile at all, which is not entirely true. As I wrote earlier every reasonably stable substance has a gas/solid equilibrium line on the phase diagram. Sure, the vapor pressure can be negligible for most practical purposes, but it still doesn't make it zero.

 

[Mr Real]

Right. That means ice disappears in the presence of brine, so as long as the brine is present there is no ice at the melting point, and to solidify brine we have to lower the temperature till the vapor pressure over the brine is equal to that over the ice, yes?
No matter how much ice evaporates, is - ever - the brine diluted infinitely, or is - always - salt concentration finite? And as long as the concentration is finite, is the vapor pressure over the brine equal, or lower to the vapor pressure over the ice (we are talking Raoult's law all the time here)?

Sorry, but I didn't understand most of what you said. The things I have understood until now is that if we place ice and brine together, the initial vapor pressure of the brine will be lower than the ice and thus condensation will happen over it( so it will be diluted to some extent and the salt concentration will decrease), and some amount of ice will evaporate, and the ice will stop evaporating as soon as both vapor pressures become equal.
So, I don't get why ice will disappear completely or that as long as brine as present, why can't ice be present too?

 

[Borek]

the ice will stop evaporating as soon as both vapor pressures become equal.

At melting point, what should be the concentration of brine for the vapor pressure over it to be identical to the vapor pressure over ice?

 

[Mr Real]

At melting point, what should be the concentration of brine for the vapor pressure over it to be identical to the vapor pressure over ice?

I don't know but I'll make a guess: zero concentration? (I don't know the reason).

 

[Borek]

No need to guessing, just use the Raoult's law - use it to calculate solvent (water) pressure over the solution, using x for a molar fraction of the salt. If x <> 0 can the pressure over the solution be identical with the pressure over the ice at melting point?

 

[Mr Real]

No need to guessing, just use the Raoult's law - use it to calculate solvent (water) pressure over the solution, using x for a molar fraction of the salt. If x <> 0 can the pressure over the solution be identical with the pressure over the ice at melting point?

I think it can, because condensation will occur over the brine raising its vapor pressure thus equalizing it with that of the ice (some amount of ice will evaporate). I don't think x>0 will stop the two vapor pressures from getting equal, am I wrong?

 

[Borek]

Stop guessing, we are a point where things can be shown precisely - and quite easily. Start with the Raoult's law:

[tex]P = P_0(1-x)[/tex]

Calculate what is x if [itex]P[/itex] equals [itex]P_0[/itex].

 

[Mr Real]

Stop guessing, we are a point where things can be shown precisely - and quite easily. Start with the Raoult's law:

[tex]P = P_0(1-x)[/tex]

Calculate what is x if [itex]P[/itex] equals [itex]P_0[/itex].

For that x should be equal to zero, that is the brine must be infinitely diluted by condensation.

 

[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).