[Big-Daddy]

On a phase diagram, the lines are "lines of equilibrium", i.e. at any point on these lines of equilibrium there is an established equilibrium between the two phases on either side of the line. This is according to ChemGuide (http://www.chemguide.co.uk/physical/phaseeqia/phasediags.html).

However the equilibrium cannot just spring up at this boundary which the line represents. Either side of it, the equilibrium must still be occurring, but less appreciably. For instance, liquids always have a liquid-vapour equilibrium, even in conditions (of T and P) not on the line in the phase diagram for that substance; only sometimes they are mostly liquid and we can call this equilibrium negligible.

Thus, what values or range of the equilibrium constant do the lines in standard phase diagrams tend to represent? e.g. looking at the phase diagram for water, http://en.wikipedia.org/wiki/File:Phase-diag2.svg, the blue line in the diagram represents the equilibrium line for the equilibrium H2O (l) <-> H2O (g), i.e. at the points of (T,P) along the line is this equilibrium constant within some appreciable range which means the phases are considered at equilibrium. What is this range of values?

[Corribus]

A rare instance where ChemGuide uses some clumsy language.

The line in a phase diagram is where the standard Gibbs energies of the two phases are exactly equal, so there is no driving force for one phase to turn into the other (or vice-versa).  So let's say you are in the liquid phase at a certain temperature.  There is always an equilibrium between the liquid and solid state, and the Gibbs energies of the two phases are temperature dependent, so you can calculate a "rate of evaporation".   

For instance, let’s look at the standard Gibbs energy change (J/mol) for water going from liquid to vaporous state (calculated from heat capacities) as a function of temperature with the pressure held constant at one atmosphere.

298.15    8558.434
303.15    7965.748
313.15    6790.625
323.15    5628.796
333.15    4479.846
343.15    3343.394
353.15    2219.084
363.15    1106.585
373.15    5.593691
383.15    -1084.17
393.15    -2162.96

You see quite clearly that there is a change in sign of the standard Gibbs energy change at about 373 K, or 100 Celcius, right where you know the boiling point is.

Now let’s take some other point, say 353.15 K.  There is still an equilibrium going on at this temperature.  The vaporization of liquid water is not thermodynamically spontaneous at this temperature, but even so there is still an equilibrium, so we know that there will always be some water present as water vapor.  So if it’s not-spontaneous, why does a puddle of water evaporate to completion even at room temperature?  Well that’s because of La Chatellier’s principle – the vapor can drift away, which continually drives the equilibrium toward vaporization, even though it’s not a spontaneous process.  Still, the standard Gibbs energy change becomes less positive as the temperature increases, which naturally predicts what we know to be the case: at higher temperature, vaporization will happen faster, because the equilibrium more and more favors the gas state.

As we go from 353.15 K and increase the temperature, at around 373.15 the sign of the Gibbs energy change switches sign from positive to negative.  Then the process of vaporization becomes spontaneous.  All this really means is that the gas state is favored over the liquid state.  Practically, nothing has really changed – there is still and equilibrium, so you will still have liquid hanging around.  Of course, boiling is a little more complex than this because of cavitation and so on, but don’t make your life too difficult.

Anyway, point is that that there is a temperature where the standard Gibbs energy change is zero, because the liquid and gas states have the same standard Gibbs energy.  At this point the gas and liquid are equally favored and there is 1 to 1 equilibrium (K = 1).  The line in the phase diagram is made by plotting out where this temperature resides for every pressure.

Make sense?

[curiouscat]

Nice analysis Corribus.

Does this say anything about rates? Say at 95 C will rate of conversion of liq to vapor be only marginally less than at 100 C?

I think not, but if so, where's the discontinuity. The smooth variation of ΔG doesn't reveal that?

Alternatively, of all the equilibria what makes the 100 C equilibrium special (such that high rate transition happens, T remains constant etc. )? The fact that ΔG = 0?  Also, in chemical reaction equilibria why isn't ΔG=0 much noteworthy as it is in boiling?

[Corribus]

I think it's hard to speak of differences in rate between boiling and evaporation because there are two different types of processes.  Any phase change system is going to be heterogeneous and thus complex.  Evaporation is primarily a surface phenomenon - evaporation can only happen at a surface, so rate is going to be significantly determined by the available surface area.  In boiling, the entropy change is favorable for spontaneous change from liquid to gas, so you have phase changes throughout the system - although cavitation and the need to form bubbles complicates any simple calculation of rates.  That is, it's possible to form metastable systems.

Liquid to gas is especially complicated because the rate is also going to be influenced by pressure.  That is, if you put water in a sealed container, the rate of evaporation is going to be time dependent, because as the water evaporates, the pressure will go up, which will slow down the rate.

[Big-Daddy]

Thanks, that post helped me. But I have a few points to clarify:

1) Is this temperature, where ΔG°=0 thus K=1, the boiling point, i.e. the point in temperature at a certain constant pressure above which K for H₂O (l)  H₂O (g) is >1 (i.e. favours gaseous H₂O)?

2) Here's the real question.

You said:

At this point the gas and liquid are equally favored and there is 1 to 1 equilibrium (K = 1).

On its own, this makes sense. It would suggest that, at this temperature and pressure, if we add some H₂O (l) now, then Q<1 whereas K=1 so the equilibrium will shift forward until Q=1 is re-established. In other words, adding liquid water or water vapour will simply result in some of that substance being converted to the other phase to meet K=1 again.

But I'm not sure then about this:

The line in a phase diagram is where the standard Gibbs energies of the two phases are exactly equal, so there is no driving force for one phase to turn into the other (or vice-versa).

If you add water in one phase or the other, without changing the temperature or pressure, then as per your above explanation we should get a driving force (ΔG, not ΔG°) pushing the reaction to reach K=1 again (this will happen at any temperature and pressure, just that K won't equal 1 except at these points, at which it is defined to equal 1).

[Corribus]

Above the boiling point, DG < 0, so K > 1.  Gas phase is favored, and this is mostly because of entropy.

On its own, this makes sense. It would suggest that, at this temperature and pressure, if we add some H2O (l) now, then Q<1 whereas K=1 so the equilibrium will shift forward until Q=1 is re-established. In other words, adding liquid water or water vapour will simply result in some of that substance being converted to the other phase to meet K=1 again.

No, because the activity of pure liquid water is unchanged regardless of how much you add.  Adding more water doesn't change Q - it doesn't impact the equilibrium, at least not directly.  Adding more water vapour, however will shift the equilibrium because as you know the activity of a gas is related to its pressure.  If the water vapor is added to a closed system, the pressure will rise, which will tend to favor condensation.

Also I again stress that vaporization is highly dependent on the available surface area, so a simple treatment of the problem using only Qs and Ks doesn't really apply.

If you add water in one phase or the other, without changing the temperature or pressure, then as per your above explanation we should get a driving force (ΔG, not ΔG°) pushing the reaction to reach K=1 again (this will happen at any temperature and pressure, just that K won't equal 1 except at these points, at which it is defined to equal 1).

Maybe my answer above helps you.  Since the activity of pure liquid water is independent of how much of it is there, adding liquid water will not impact the equilibrium because Q (abd hence DG) will be unchanged*.  Adding water vapor will increase the pressure, which will shift the equilibrium because Q will be changed.  There will then be a driving force to reform equilibrium.

One last point here, related to the *, is that unlike in many simple chemical reactions, here the pressure dependence cannot be neglected.  In a closed system, adding some liquid will reduce the volume available for the vaporized gas, so technically the equilibrium will be shifted - although not for the reason you've mentioned.  Likewise, adding water vapor shifts the equilibrium due to a change in the activity of the gas, which is related to the pressure change.

So you see the system is very complex and I think you have to approach it differently than the way you approach reaction equilibria.

[Big-Daddy]

No, because the activity of pure liquid water is unchanged regardless of how much you add.  Adding more water doesn't change Q - it doesn't impact the equilibrium, at least not directly.  Adding more water vapour, however will shift the equilibrium because as you know the activity of a gas is related to its pressure.  If the water vapor is added to a closed system, the pressure will rise, which will tend to favor condensation.

I'm not sure if this explains things properly. Take H₂O at temperature 403.15 K and standard pressure of 1 bar. The liquid-gas equilibrium H₂O (l)  H₂O (g) has K=p[H₂O (g)]/a[H₂O (l)] which you say we can rewrite K=p[H₂O (g)] (I would have thought the assumption that pure liquids are at unity activity no longer holds for phase equilibria?), where K>>1 at this T and P so that most of the water is gaseous. Surely if we add H₂O (l), there will be a driving force for its conversion to H₂O (g)? It will not just sit there - which indicates that Q has been reduced in value from K to something well below K, meaning the forward reaction occurs more than the backward reaction, and thus that Q has been affected by adding liquid water. 

One last point here, related to the *, is that unlike in many simple chemical reactions, here the pressure dependence cannot be neglected.  In a closed system, adding some liquid will reduce the volume available for the vaporized gas, so technically the equilibrium will be shifted - although not for the reason you've mentioned.  Likewise, adding water vapor shifts the equilibrium due to a change in the activity of the gas, which is related to the pressure change.

So you see the system is very complex and I think you have to approach it differently than the way you approach reaction equilibria.

I don't mind dependence on pressure - as I am aware, all equilibrium constants have at least a little bit of dependence on pressure, as well as on temperature.

[Corribus]

BD,
You are overthinking it.  Though, I think I may have misled you.

You start with

Take H2O at temperature 403.15 K

At 1 atmosphere, how do you have water at this temperature?  It's not possible.  What I wrote above (providing delta G values above the boiling point) was something of an abstraction.  Once the temperature reaches the boiling point, I don't view it as a simple liquid-gas equilibrium any longer (in an open container) because there is enough energy to overcome atmospheric pressure and induce bubble formation.  I think it's a more complex problem at that point and you can't treat it the same way.  Under 1 atmosphere of pressure, it's not possible to heat water higher than the boiling point (ignoring metastable states).  In a closed container, of course, you can get water this hot, because the total pressure in the container will rise with the temperature - but again this is a completely different problem because the external temperature is also changing.

Let's stick to a temperature below the boiling point.

Put water in a container at 50 Celsius.  The vapor pressure is 12.3 kPa.  Now double the amount of water at the same temperature.  Does the vapor pressure change?  No.  This is because the activity of the "pure" liquid water is approximately 1 and therefore the equilibrium constant (and Q) doesn't dependent on the amount of water present.  The vapor pressure is essentially an intensive property. 

Now, at high pressures this approximation of the activity of liquid water being 1 begins to weaken, because the standard state is defined for 1 atmosphere (just as the approximation begins to weaken in concentrated solutions, because the water can no longer be approximated as "pure").  In these cases you would have to do something else... use the liquid fugacity or something.  I don't know - chemical engineers probably would have to worry about that kind of stuff so you'll have to ask one of them.

Other than that's it's a pretty simple equilibrium problem and the standard approximations apply:

http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Virtual%3A_Phase_Changes/Vapor_Pressure (mind the silly typo in the first equation - they aren't talking about hydrogen gas)
http://chemed.chem.wisc.edu/chempaths/GenChem-Textbook/The-Equilibrium-Constant-in-Terms-of-Pressure-965.html

Anyway, phase equilibria are complicated and not always intuitive and it'll make your head hurt if you think too much about them.  I know it does mine.

[curiouscat]

  In these cases you would have to do something else... use the liquid fugacity or something.  I don't know - chemical engineers probably would have to worry about that kind of stuff so you'll have to ask one of them.

Ha! We simply use a Moeller chart or a steam table and get on with life. 

[Big-Daddy]

At 1 atmosphere, how do you have water at this temperature?  It's not possible.  What I wrote above (providing delta G values above the boiling point) was something of an abstraction.  Once the temperature reaches the boiling point, I don't view it as a simple liquid-gas equilibrium any longer (in an open container) because there is enough energy to overcome atmospheric pressure and induce bubble formation.  I think it's a more complex problem at that point and you can't treat it the same way.  Under 1 atmosphere of pressure, it's not possible to heat water higher than the boiling point (ignoring metastable states).  In a closed container, of course, you can get water this hot, because the total pressure in the container will rise with the temperature - but again this is a completely different problem because the external temperature is also changing.

Thanks, the rest of your post is clear, so this is the bit I'm honing in on.

I'm not sure why it should make a difference to the phase of the water itself whether the container is open or not, so long as the water's temperature can be governed (adiabatically, i.e. we assume we can set a certain temperature for the water and keep it there). Phase diagrams for water all show water as being gaseous at 400 K and 1 Pa (example: http://www.lsbu.ac.uk/water/phase.html) which would suggest these conditions are possible in theory; how then would the liquid water be spontaneously converted to vapour, which it will, to maintain the equilibrium (suggesting Q has been affected)? Perhaps the approximation of the activity of liquid water being unity becomes substantially weaker as the temperature rises as well as the pressure?

[Big-Daddy]

i.e. if you add water in one phase or the other, at any T or P, without changing the temperature or pressure, then because water's activity cannot be approximated as 1 for phase equilibria (impression I'm getting - if not, then please explain what exactly makes Q change as you add more water), then the activity of liquid water changes when you add the liquid water so Q changes and this causes the reaction to move to produce more water vapour (reducing the activity of the liquid water again, as the quantity of liquid water is reduced through conversion to gaseous water) to reach the same K as specified at that particular T,P for this phase equilibrium.

[Enthalpy]

[...] the equilibrium cannot just spring up at this boundary which the line represents. Either side of it, the equilibrium must still be occurring, but less appreciably. For instance, liquids always have a liquid-vapour equilibrium, even in conditions (of T and P) not on the line in the phase diagram for that substance; only sometimes they are mostly liquid and we can call this equilibrium negligible.

No. The equilibrium is a line. One pressure for each temperature, one temperature for each pressure - over the range where the equilibrium exists.

[Big-Daddy]

No. The equilibrium is a line. One pressure for each temperature, one temperature for each pressure - over the range where the equilibrium exists.

This doesn't agree with what Corribus was saying above ...

Are you saying that the equilibrium is there along the line, and then simply vanishes off the line? Surely this cannot be; the H₂O (l) H₂O (g) must to some extent be reversible because all reactions are. And furthermore if equilibrium only exists on the line then this fails to explain the presence of vapour pressure at points of T,P not on the line.

[Big-Daddy]

Can we come to some kind of agreement on this? I don't feel comfortable moving on before I've established basically whether or not there is an equilibrium on the line (with the line representing T and P where K=1) and if not, how these problems I mentioned above are explained.

[curiouscat]

I think the rest of us have an agreement.

Equilibrium is indeed on the line. @Enthapy seems absolutely right. Where does @Corribus say otherwise?