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.
O (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 H2
O (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 H2
O (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)?