Why does adding solute particles decrease the freezing point?

[Mr Real]

When solute particles are added to a solvent boiling point increases, I am quite clear with its explanation (in terms of vapour pressure) but when it comes to freezing point, why does it decrease? What does it have to do with vapour pressure? My textbook says it's because liquid and solid state of the solvent exist in equilibrium at freezing point and there is a vapor pressure due to this, but how can a liquid or solid exert vapour pressure?
Also, why doesn't increase in boiling point depend upon the nature of solute particles, I would imagine their size, volume and other properties would affect this but apparently they don't?

 

[Borek]

how can a liquid or solid exert vapour pressure?

Does the liquid dry without a boiling?

Have you heard about sublimation?

 

[Mr Real]

Have you heard about sublimation?

But sublimation happens only for a special category of solids: volatile solids. But I am not talking specifically about volatile solids, I want to know what is the cause of vapour pressure of any solution at its freezing point and why is a solution's freezing point lower than that of the original solvent's?

 

[Borek]

sublimation happens only for a special category of solids

No, it happens for every solid. Every phase diagram contains a solid/gas equilibrium line.

If you have two phases at a given temperature, one with a higher saturated vapor pressure above, and the other with a lower saturated vapor pressure above, what will happen?

 

[Mr Real]

If you have two phases at a given temperature, one with a higher saturated vapor pressure above, and the other with a lower saturated vapor pressure above, what will happen?

I don't know what will happen then because I didn't know there were 2 vapor pressures involved in a phase change. But then when a liquid boils, I think what happens is some of the liquid evaporates and some of its vapour condenses and there's only one vapor pressure involved, that of the vaporized liquid. I have no clue regarding the freezing point.

 

[Borek]

I don't know what will happen then because I didn't know there were 2 vapor pressures involved in a phase change.

You are on the right track - there can't be two vapor pressures. That means if there are two phases with two different vapor pressures, one of the phases is unstable - it evaporates, and at the same time the vapors condensates producing the other phase.

This is enough information to explain the depression in the freezing point.

 

[Mr Real]

So, please

You are on the right track - there can't be two vapor pressures. That means if there are two phases with two different vapor pressures, one of the phases is unstable - it evaporates, and at the same time the vapors condensates producing the other phase.

This is enough information to explain the depression in the freezing point.

Correct me when I'm wrong. When a pure solvent freezes, there exists an equilibrium between its liquid-vapour state change(and no equilibrium between its solid-vapour state change but we are talking about freezing here, what about a solid-liquid change?) and then when a solute is dissolved in it, what happens? You have not mentioned about this component, only about two phases (by which I understand you mean state).

 

[Borek]

an equilibrium between its liquid-vapour state change

No idea what you mean by that. There can be an equilibrium between two different coexisiting phases, but "equilibrium between state change" doesn't mean anything.

Phase is a concept more general than the state of matter - for example solid phosphorus is a state of matter (just like liquid and gas are), but this solid can be composed of several phases, that's what allotropy is about.

Let's start with something simpler. Imagine you have a large, closed container. Inside there are two bakers, one containing 20% brine, the other containing 5% brine. What will happen and why? Hint: is the saturated vapor pressure over the brine the same in both bakers? Can a single container contain two different vapor pressures at the same time?

 

[Mr Real]

What will happen and why? Hint: is the saturated vapor pressure over the brine the same in both bakers? Can a single container contain two different vapor pressures at the same time?

According to me, No, the saturated vapor pressure will be different in both the cases: it will be higher in the 5% brine. Regarding the second hint, I think there will only be one vapor pressure but whether it will be the smaller, the larger or the sum of the 2 individual vapor pressures I don't know (though I think it should be the sum) ?

 

[Borek]

You are right there can be only one pressure, and you are right when the concentrations are different there are two different pressures. Can you think of a process that will eliminate the problem?

 

[Mr Real]

Can you think of a process that will eliminate the problem?

I can't think of a process to eliminate the problem other than mixing the 2 solutions.

 

[DrClaude]

Have a look at this page: https://chem.libretexts.org/Textboo...Point_Elevation_and_Freezing_Point_Depression

I like the last figure in particular. You can see how the change in Gibbs free energy going from a pure solvent to a solution changes both the freezing and boiling points.

 

[Borek]

I can't think of a process to eliminate the problem other than mixing the 2 solutions.

What about evaporation and condensation?

 

[Mr Real]

What about evaporation and condensation?

So, are you saying that when we keep the two breakers then without any intervention needed, naturally the processes of evaporation and condensation will equalise the two vapor pressures, to maybe either of the individual vapor pressures or the sum of these pressures?

 

[Borek]

So, are you saying that when we keep the two breakers then without any intervention needed, naturally the processes of evaporation and condensation will equalise the two vapor pressures

Exactly.

to maybe either of the individual vapor pressures or the sum of these pressures?

None of these. Vapor pressure depends on the concentration of the solution, doesn't it? When will the vapor pressure be identical over the content of each beaker?

 

[Mr Real]

None of these. Vapor pressure depends on the concentration of the solution, doesn't it? When will the vapor pressure be identical over the content of each beaker?

I think that will be when the concentrations are equal in both the beakers (but in this case they are both different so what will happen here?)

 

[Borek]

I think that will be when the concentrations are equal in both the beakers

Right.

but in this case they are both different so what will happen here?

Think about evaporation and condensation.

 

[Mr Real]

Think about evaporation and condensation.

Okay, the vapor pressures will be equal over both of them but my question( this is out of curiosity ) is this: If we had put each of the beakers alone in the container before putting them together then they would have had different vapor pressures, right ( say, V1 & V2 )? Now, when we put both of them inside together, will this vapor pressure be the sum of V1 and V2,will be equal to either of them or will have some other relation to them depending on other factors?

 

[Borek]

Initially it would equal the greater of the two. However, what happens when the vapor pressure is higher than the equilibrium pressure (in other words: what happens when the vapor is supersaturated)?

 

[Mr Real]

Initially it would equal the greater of the two. However, what happens when the vapor pressure is higher than the equilibrium pressure (in other words: what happens when the vapor is supersaturated)?

Condensation would happen and thus we would obtain the saturated vapor pressure.

 

[Borek]

Where would the condensation take place? In both solutions to same extent, or preferentially in one of them?

 

[Mr Real]

Where would the condensation take place? In both solutions to same extent, or preferentially in one of them?

According to me, condensation would preferentially take place in the beaker having 20% brine.

 

[Borek]

Perfect. It has a lower vapor pressure, so from its "point of view" the vapor is supersaturated and will condense there. Now, what will happen to the concentrations of brine in both beakers?

 

[Mr Real]

Perfect. It has a lower vapor pressure, so from its "point of view" the vapor is supersaturated and will condense there. Now, what will happen to the concentrations of brine in both beakers?

The concentrations in both the beakers will become equal ( But I don't know if it'd be 5 or the mid value of 5 & 20 or some other value)

 

[Borek]

Some mid value - depends on the exact amount of both solutions. Actually it won't be different from the concentration you would get by combining both solutions.

OK, now that we have established that water moves from a source that has a higher vapor pressure to the place where the equilibrium vapor pressure is lower, can you apply the same principle to another container containing a piece of ice and a beaker with a brine solution? Do you know the Raoult's law?

 

[Mr Real]

OK, now that we have established that water moves from a source that has a higher vapor pressure to the place where the equilibrium vapor pressure is lower, can you apply the same principle to another container containing a piece of ice and a beaker with a brine solution?

Here, I think condensation would take place in the piece of ice (since V.P. of ice,solid < V.P. of brine solution,liquid). But it seems wrong somehow.

Do you know the Raoult's law?

Yes, I do know about the Raoult's law.

 

[Borek]

Here, I think condensation would take place in the piece of ice (since V.P. of ice,solid < V.P. of brine solution,liquid).

Let's start at the melting point. Both phases (liquid and solid) coexist, don't they? Could they coexist if their vapor pressures were different?

 

[Mr Real]

Let's start at the melting point. Both phases (liquid and solid) coexist, don't they? Could they coexist if their vapor pressures were different?

No, I don't think that solid and liquid states of the same substance can exist in equilibrium if their vapor pressures are different( because the vapor pressure will be of that same substance and it can't be different for both the states).

 

[Borek]

OK, so the obvious conclusion is that at the melting point liquid and solid can coexist BECAUSE their vapor pressures are identical.

What about a solid and brine at the melting point of water? Is the vapor pressure over brine the same as over the ice?

 

[Mr Real]

What about a solid and brine at the melting point of water? Is the vapor pressure over brine the same as over the ice?

As they are placed in the same container/system, I think that the vapor pressure will be equal over the brine as well as the ice.

 

[Borek]

As they are placed in the same container/system, I think that the vapor pressure will be equal over the brine as well as the ice.

Yes, but is this an equilibrium vapor pressure?

 

[Mr Real]

Yes, but is this an equilibrium vapor pressure?

If you mean if it'd be the same as the vapor pressure if the ice or the brine had been placed alone in the container, then I think it wouldn't. But if you didn't mean that, then I can't think why it isn't an equilibrium vapor pressure.

 

[Borek]

If you mean if it'd be the same as the vapor pressure if the ice or the brine had been placed alone in the container, then I think it wouldn't.

So now you wrote the pressures over brine and solid are different, but before you wrote they are identical:

I think that the vapor pressure will be equal over the brine as well as the ice.

We have already established, that if you have two samples with different equilibrium vapor pressures in the same container, mass will be transported from one sample to another till the system reaches the equilibrium (whatever it means).

 

[Mr Real]

So now you wrote the pressures over brine and solid are different, but before you wrote they are identical:

No, I meant that if we had placed either ice(with VP. v1) or brine(with V.P. v2 ) alone in the container then these would be different from the equilibrium vapor pressure we'd get if we placed both of them together.

 

[Borek]

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.

 

[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:

P = P_0(1-x)

Calculate what is x if P equals P_0.

 

[Mr Real]

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

P = P_0(1-x)

Calculate what is x if P equals P_0.

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