Why does adding solute particles decrease the freezing point?

[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.

 

[Mr Real]

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

Yeah, I can see that but I was talking about the absolute value of the difference in vapor pressures.

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).

Okay, I didn't notice that, I thought the minimum was at 273 K in the plot.

So, if we move from 273 down, the difference gets lower till it hits zero, it doesn't grow, at least initially.

Yes, I agree it doesn't grow initially.

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)

Well, I've been talking about the absolute value for all this time. The absolute value of the difference in vapor pressures is increasing whether we go to lower/higher temperatures (except when it decreases initially when we reduce temperature from 273 K to 271.3 K, as you said).

 

[Borek]

Okay, I didn't notice that, I thought the minimum was at 273 K in the plot.

How could it me minimum at 273, when the difference of zero makes it possible for both phases - ice and solution - to coexist? And we are talking about the freezing point depression, so the zero (coexistence of liquid and solid) must be at some other, lower temperature?

except when it decreases initially when we reduce temperature from 273 K to 271.3 K

So you have missed the only part of the plot that matters for the whole discussion about the freezing point depression. Again, you are concentrating on things that are irrelevant to the discussion making me wonder if you understood the important part

 

[Mr Real]

So you have missed the only part of the plot that matters for the whole discussion about the freezing point depression.

Sorry, but I didn't know that and it was kind of a minute detail in the graph.

Again, you are concentrating on things that are irrelevant to the discussion

You asked me to follow the plot, so I did and listed all my observations.

making me wonder if you understood the important part

I have understood the important part.

 

[Mr Real]

Borek, can you please answer my doubt? Was I right when I said in post #87 that the 2 vapor pressures can become equal only if the decrease in vapor pressure is less for the solution (brine in our case) than for the solid state of the solvent (ice in our case)? And if I was right then does this hold for every solvent?
Thanks
Mr R

 

[jim mcnamara]

Mr. Real - it seems to me that you have some kind of issue understanding what @Borek has done for you. It is a great explanation! If you cannot focus clearly enough to come to an understanding of the material, then some of us with less patience than Borek, like me, will simply close the thread.

Thanks in advance for your help. We want you to learn, not to parasitize someone's time and effort when you do not appear to make the effort yourself.

 

[snorkack]

Borek, can you please answer my doubt? Was I right when I said in post #87 that the 2 vapor pressures can become equal only if the decrease in vapor pressure is less for the solution (brine in our case) than for the solid state of the solvent (ice in our case)?

Strictly speaking, were it otherwise then the vapour pressures could become equal on heating rather than cooling.

And if I was right then does this hold for every solvent?

Well, helium (both isotopes) is unique for having the property to freeze on heating.

 

[Mr Real]

Mr. Real - it seems to me that you have some kind of issue understanding what @Borek has done for you. It is a great explanation! If you cannot focus clearly enough to come to an understanding of the material, then some of us with less patience than Borek, like me, will simply close the thread.

Thanks in advance for your help. We want you to learn, not to parasitize someone's time and effort when you do not appear to make the effort yourself.

Sorry, Borek if I came across that way. I am not deliberately trying to waste your time by behaving stupidly, I am genuinely not clear with the things I asked.
jim mcnamara, I do recognise Borek's helping attitude and patience in trying to clear my doubts, and because of that most of my doubts have been cleared. Some doubts(which I mentioned in the last post) remain and I've really tried to answer this myself and in that regard have asked Borek's opinion on whether I'm right or not. I wholeheartedly agree his explanation has been really great, with analogies I've not even found in my textbook!
Thanks so much !
(and sorry @Borek )

 

[Borek]

Was I right when I said in post #87 that the 2 vapor pressures can become equal only if the decrease in vapor pressure is less for the solution (brine in our case) than for the solid state of the solvent (ice in our case)?

Isn't it clear from the plot? And as snorkack wrote, it is quite easy to realize it can't be other way, as it won't lead to the existence of the melting point at all.

And if I was right then does this hold for every solvent?

https://www.physicsforums.com/threads/depression-of-freezing-point.915216/page-5#post-5776336

(and don't worry about exotic cases, they definitely make nature more interesting, but they are rare enough to be ignored at the level you are now)

 

[snorkack]

And as snorkack wrote, it is quite easy to realize it can't be other way, as it won't lead to the existence of the melting point at all.

(and don't worry about exotic cases, they definitely make nature more interesting, but they are rare enough to be ignored at the level you are now)

Or more precisely, being otherwise would lead to the exotic case of melting on cooling rather than heating.

 

[Mr Real]

Thank you @Borek and @snorkack for clearing my doubts.

Mr R

 

[jim mcnamara]

Thread came to a good ending. Thank you.