[curiouscat]

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 ...

Yes it does.

Are you saying that the equilibrium is there along the line, and then simply vanishes off the line?

Yes.

Surely this cannot be; the H₂O (l) H₂O (g) must to some extent be reversible because all reactions are.

Ok. So?

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.

Nope. Vapor can be present. Just not at equilibrium. When things equilibriate the parameters will again fall on the P-T curve.

[Big-Daddy]

I think the rest of us have an agreement.

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

Here:

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,

And in the general argument I read into that post, it suggests there is always an equilibrium (which is temperature and pressure-dependent) between any two phases at any values of temperature and pressure. And as Corribus goes on to point out:

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.

Which says that the line is plotted from the points where the equilibrium constant is equal to 1 - the equilibrium constant being implied to exist at all points of T, P.

[Big-Daddy]

Yes it does.

How? Read my post above ...

Surely this cannot be; the H₂O (l) H₂O (g) must to some extent be reversible because all reactions are.

Ok. So?

So H₂O (l)  H₂O (g) exists at every point on the P, T curve. By the very definition of this reaction being reversible.

Nope. Vapor can be present. Just not at equilibrium. When things equilibriate the parameters will again fall on the P-T curve.

Ah now I am beginning to understand. But I'm not sure it makes sense - if I'm at a point of P,T not on the line, and I fix the conditions there, there is no vapour pressure? (Either the substance is all vapour, or there is no vapour.)

So the reversible reaction H₂O (l)  H₂O (g) is present under all conditions. Somehow, under conditions of P,T not on the curve, the reaction proceeds in one direction more than in the other, and continues to do so until the P,T conditions are brought back to those on the line, at which point K=1 for the reversible reaction so there will be the same amount (in terms of activity/partial pressure) of the two phases present. This suggests that the equilibrium constant for H₂O (l)  H₂O (g) is 1 but this only becomes relevant when at conditions on the line because off the line, something (dependent on pressure and temperature) ensures that the rate in a certain direction is always higher than the rate in the other. Is this what you're saying? Even then, this requires explanation.

Why does the reaction proceed in one direction more than the other, regardless of the fact that the activity of one side (the reactant in the favoured direction) is decreasing almost to 0? (Because rate in a certain direction is normally a function of the activities on the reacting side of that direction.) And even if the activity of the pure liquid isn't seen as changing, the partial pressure of the vapour is definitely changing, and even then the question remains, what forces the reaction in particular directions and then does not allow them to reach equilibrium under all conditions? (Equilibrium for most processes is achieved over time regardless of the temperature conditions I thought)

What causes the P,T to change as the reaction proceeds in a certain direction, to level off at points on the line when equilibrium can actually be reached?

And also, will there be no vapour and thus no vapour pressure present over a liquid if I hold the conditions to a P, T point not on the phase line (i.e. inside the liquid region)?

[curiouscat]

When you have this equilibrium how many phases and components are there.

Ergo how many degrees of freedom.

[Big-Daddy]

When you have this equilibrium how many phases and components are there.

Ergo how many degrees of freedom.

2 phases, 1 component. I don't know how to calculate degrees of freedom, but from a quick online search for degrees of freedom I can't conceivably draw links which would help me answer any of the questions above.

If you explain whether the understanding I suggested in the post above is correct or not that would be a start:

So the reversible reaction H₂O (l)  H₂O (g) is present under all conditions. Somehow, under conditions of P,T not on the curve, the reaction proceeds in one direction more than in the other, and continues to do so until the P,T conditions are brought back to those on the line, at which point K=1 for the reversible reaction so there will be the same amount (in terms of activity/partial pressure) of the two phases present. This suggests that the equilibrium constant for H₂O (l)  H₂O (g) is 1 but this only becomes relevant when at conditions on the line because off the line, something (dependent on pressure and temperature) ensures that the rate in a certain direction is always higher than the rate in the other. Is this what you're saying?

[curiouscat]

2 phases, 1 component. I don't know how to calculate degrees of freedom, but from a quick online search for degrees of freedom I can't conceivably draw links which would help me answer any of the questions above.

P + F = C + 2

[curiouscat]

Your basic misconception here seems that you can take a two phase system, specify both T and P arbitrarily and then get it to equilibrium under those P and T conditions.

Won't happen.

[Big-Daddy]

Your basic misconception here seems that you can take a two phase system, specify both T and P arbitrarily and then get it to equilibrium under those P and T conditions.

Won't happen.

I know, this is what I said in the post I have referred to twice before:

So the reversible reaction H₂O (l)  H₂O (g) is present under all conditions. Somehow, under conditions of P,T not on the curve, the reaction proceeds in one direction more than in the other, and continues to do so until the P,T conditions are brought back to those on the line, at which point K=1 for the reversible reaction so there will be the same amount (in terms of activity/partial pressure) of the two phases present. This suggests that the equilibrium constant for H₂O (l)  H₂O (g) is 1 but this only becomes relevant when at conditions on the line because off the line, something (dependent on pressure and temperature) ensures that the rate in a certain direction is always higher than the rate in the other. Is this what you're saying?

Several questions remain. Let's start with the main one:

Why does the reaction proceed in one direction more than the other, regardless of the fact that the activity of one side (the reactant in the favoured direction) is decreasing almost to 0? (Because rate in a certain direction is normally a function of the activities on the reacting side of that direction.) And even if the activity of the pure liquid isn't seen as changing (BTW: not at all valid once the liquid is entirely converted to gas, activity is 0 then!), the partial pressure of the vapour is definitely changing. I.E. what force causes the reaction to keep moving forward in a certain direction, without equilibrium ever being reached (at values of T and P not on the line).

[curiouscat]

Why does the reaction proceed in one direction more than the other, regardless of the fact that the activity of one side (the reactant in the favoured direction) is decreasing almost to 0?

Why? I can't see any activity going down to zero here.

[Big-Daddy]

Why? I can't see any activity going down to zero here.

At any point in P and T the reversible reaction H₂O (l)  H₂O (g) is occurring. But let's say we're at a point of P, T squarely inside the gaseous region, and we're going to hold things at these conditions. Then the forward rate for the reaction overpowers the backward rate - for a reason you have not explained to me. Most importantly, the forward rate stays fairly constant because the activity of the water stays fairly constant (and the conditions are held constant); but the backward rate should be increasing, as the vapour increases in partial pressure, until it can match the backward rate and equilibrium is reached. Where's the problem with the logic there?

And still there remains the question, what are the forward and backward rates functions of, that one continues to dominate the other regardless of how the activity or partial pressure on one side changes?

[curiouscat]

as the vapour increases in partial pressure,

How is that consistent with your "we're at a point of P, T squarely inside the gaseous region, and we're going to hold things at these conditions"

Either the pressure is constant or it isn't.

You should double check your hypotheticals for consistency (I hope I don't have to eat crow!).

[Big-Daddy]

How is that consistent with your "we're at a point of P, T squarely inside the gaseous region, and we're going to hold things at these conditions"

Either the pressure is constant or it isn't.

You should double check your hypotheticals for consistency (I hope I don't have to eat crow!).

Valid point. However, maybe the volume of the container is being changed to keep pressure constant? What then.

I only pose this because I want to see what happens if P and T are constant (and yet, obviously, we need to get a gas being produced more than a liquid if we're at a (P,T) point in the gaseous section).

[curiouscat]

Valid point. However, maybe the volume of the container is being changed to keep pressure constant? What then.

What then? Pressure is constant. There fails your argument which said: " but the backward rate should be increasing, as the vapour increases in partial pressure,"

No increase in Partial Pressure. End of story.

I only pose this because I want to see what happens if P and T are constant (and yet, obviously, we need to get a gas being produced more than a liquid if we're at a (P,T) point in the gaseous section).

Yes. That happens.  We end up with all gas at P,T. Where's the contradiction?

[Big-Daddy]

What then? Pressure is constant. There fails your argument which said: " but the backward rate should be increasing, as the vapour increases in partial pressure,"

No increase in Partial Pressure. End of story.

Point taken. I feel like I'm understanding more now.

Two questions to check then: 1) so when the equilibrium is reached, is the constant equal to 1 (so that when the equilibrium is reached and P levels off with time, there are equal activities/fugacities of the substances present - bearing in mind of course that activity of a liquid or solid is not necessarily proportional to number of moles of that substance)?

2) How does vapour pressure come about at (P,T) points other than those on the line. e.g. we live in 298 K at 10⁵ Pa pressure, a point on the diagram in which water lies squarely within the liquid form, so how does water vapour exist under these conditions?

[curiouscat]

2) How does vapour pressure come about at (P,T) points other than those on the line. e.g. we live in 298 K at 10⁵ Pa pressure, a point on the diagram in which water lies squarely within the liquid form, so how does water vapour exist under these conditions?

If you are referring to everyday water vapor it is mixed with air. The partial pressure of water vapor in that air-water mix will fall on that P-T line.

OTOH, if it is a non-equilibrium situation none of this applies. It can pretty much do whatever it wants.