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Phase diagrams and equilibria

  1. Jun 18, 2013 #1
    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. The equilibrium constant is a function of temperature and pressure, and the various other thermodynamic factors (e.g. specific volume, specific enthalpy, or specific entropy) you may plot phase diagrams for.

    However the equilibrium does not 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 usually have a liquid-vapour equilibrium which I presume would fall under this category too.

    Thus, my first question is, 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?
     
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  3. Jun 19, 2013 #2

    DrDu

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    I don't quite get what is your question, but I think you might have some misconception here.
    Equilibrium of phases does not mean that there exists an equilibrium constant. Instead of there existing some relation between the amounts of the two phases, the two phases can coexist in any amount along that line.
     
  4. Jun 19, 2013 #3
    And the two phases cannot coexist at all outside that line? Example: Let's say that at 298.15 K and 10^5 Pa pressure, according to phase diagrams, H2O is a liquid and this does not lie on a line of equilibrium. Then, if we introduce some gaseous H2O or solid H2O, will 1-0% of it spontaneously convert to liquid H2O? And yet, if we were on a line of equilibrium and we introduced some gaseous or solid H2O, 0% would spontaneously convert to liquid (there is no equilibrium; phases simply do not interconvert along the line of equilibrium) and none of the liquid would spontaneously convert to either of these. This seems odd to me, as I thought all chemical and physical changes should properly be equilibria, even if some are sometimes assumed to be irreversible. It also raises the problem of, if all the gas is supposed to convert spontaneously to liquid when the main form is a liquid, why do we have liquid-vapour equilibria?

    If this is true, then is the only purpose of the line of equilibrium to show the conditions under which both phases can be present? Not the ratio of one phase to another, or which phase is dominant - these things simply don't exist? (i.e. at any point on the line, the effect is the same: at these points, both phases can coexist, though neither phase will spontaneously convert to any of the other phase; rather, immediately outside the line, all phases will spontaneously convert entirely to the other).
     
  5. Jun 21, 2013 #4

    DrDu

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    That's all correct. Note that in the Phase diagram which you have in mind, p is held constant.
    You get some kind of equilibrium consideration with a well defined ratio of the phases once you hold V and n constant. I.e. for a given volume and amount of substance, neither the liquid nor the gas can fill all of the volume and be stable at the same time. Rather some of the liquid will vaporize until the pressure of the gas reaches the phase equilibrium.
     
  6. Jul 5, 2013 #5
    Allow me to throw a wrench into the gears here. I can see some substance to BigDaddy's thinking. Two phases can coexist outside of the equilibrium line on the phase diagram, but not for long. This can be due to surface tension of the liquid, lack of nucleation sites for boiling or something similar. For example, superheated liquid water with water vapor, or subcooled liquid water with ice. The fugacity of each phase will be different so they will NOT be in equilibrium. There will be some kinetic rate of change to an equilibrium amount of each phase, with a corresponding rate constant as with a true chemical reaction that gets modeled by a rate law. Also there will be an equilibrium ratio of amounts of each phase, this is the equilibrium constant most people think of- this is the state of the system after gibbs free energy is locally minimized. There is always, always, always confusion between the two (equilibrium constant and rate constant). but this is really not the conventional way to consider phase changes.

    The suggested scenario of adding vapor at a temperature above boiling point to a liquid at a temperature and pressure below its boiling point will either condense some of the vapor or evaporate some of the liquid until the equilibrium line at that particular pressure is reached. This is actually sort of difficult to imagine as an isobaric process, but I won't attempt to go the ideal route of a constant volume process. The point is that initially there will be liquid and vapor of low and high temperature respectively. Imagine for the sake of simplicity that they reach thermal equilibrium quickly before the phase change process starts- thus they will both reach the saturation temperature and either vaporization or condensation starts. In this case, you will have either superheated liquid or subcooled vapor, causing a time dependent mass transfer between phases (potentially described by a rate constant but again that's not the traditional way). The two phases will momentarily be coexisting off of the equilibrium line, but will eventually reach an equilibrium ratio of vapor/liquid. This could be described by an equilibrium constant, but again, it's not done this way. When the liq/vap ratio becomes constant, they will be on the equilibrium line of the phase diagram.
     
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