Update: the American Chemical Society (which made the question) has responded to some complaints regarding this question. I don't really understand their answer (what does free energy have to do with anything?) but maybe someone else does:

At that temperature and pressure, there must certainly be NH3(g). But solid and liquid phases need not be present even if the system is at equilibrium at that temperature. Thus, a sample of NH3(g) prepared at 195.4 K and 0.0606 bar need not condense into solid or liquid; it could remain gaseous in perpetuity. Likewise, while solid and liquid can coexist at this temperature and pressure, if you start with only liquid, it need not freeze into any solid (or vice versa).

B and C are different from A because for gases, free energy depends on pressure, whereas for pure solids and liquids it does not. Thus, if solid or liquid were present but the gas pressure was not the vapor pressure, then some of the condensed phase would evaporate (or some gas would condense) to change the free energy of the gas until it was equal to that of the solid or liquid. Conversely, if only gas were present (at the correct pressure), it would not condense (in a rigid container), because doing so would lower its free energy below that of the solid or liquid that formed.

The point about not being in equilibrium has some merit, in the sense of the philosophical conundrum of “If a only one side of a chemical reaction is present, can the reaction be said to be at equilibrium?” (For example, the ice-water equilibrium at 25 °C.) The point is arguable both ways, in my view, but note that the problem said that the sample was at equilibrium, not any particular chemical reaction. The only reasonable way I can see to interpret equilibrium in this context is that the composition of the sample would be indefinitely stable. As described above, at the pressure and temperature of the triple point, that could be true of a pure gas (or a mixture of gas and liquid only, or of gas and solid only, or of course of a mixture of solid, liquid, and gas).

I think this is actually an outstanding question for assessing students’ understanding of the nature of phase equilibria—for example, the fundamental differences between gases (where pressure affects their free energy) and solids and liquids (which have fixed free energy, independent of their amount, at a given temperature). And as our discussion indicates, it is an excellent question for causing students to stop and think, either during the exam or afterwards. On a competition (not assessment) exam, especially one like the USNCO that is taken by high-achieving students, I think that it is crucial to have at least a few of these questions on each exam, to not just assess their current understanding but to actually stimulate a deepening of that understanding.

A recent United States National Chemistry Olympiad question asked the following:

Which statement best describes a sample of ammonia at equilibrium at its triple point (195.4 K, 0.0606 bar)?

A) Gaseous ammonia must be present, and solid or liquid ammonia may be present.
B) Liquid ammonia must be present, and solid or gaseous ammonia may be present.
C) Solid ammonia must be present, and liquid or gaseous ammonia may be present.
D) Gaseous, liquid, and solid ammonia must all be present.

I thought the answer was D; since it's the triple point, all three states are in equilibrium, so they should all be present to some degree. However, the answer was actually A. I think I get the basic reason why: looking at the triple point on a phase diagram (phase diagram of ammonia shown below), the large portion below the triple point is purely gas, but above the triple point, it's split between solid and liquid. But if that is the reasoning, I don't quite agree with it. Since it's the triple point, all three states must be in equilibrium, and as far as I know, you can't have an equilibrium constant of zero or infinity (I know equilibrium constants might not apply to solids but I'd imagine the concept is the same). If two or more substances could be in equilibrium with each other without one of them being present, you could point to a sample of lead and say it's in equilibrium with gold. Could someone please shed some light on this?

Phase Diagram of Ammonia

  • $\begingroup$ As a guess, it might be considering the limiting case where there is exactly enough vapor to fill a container to 0.0606 bar, or at the instant of time, in an open system, where all liquid and solid have evaporated. But that guess does not leave me satisfied. $\endgroup$ Commented Apr 4 at 2:09
  • $\begingroup$ Can you have pure liquid water or pure ice at the normal freezing point of water? Or do you have to have liquid and solid? As you need some nuclei for the ice to form and it is possible to have supercooled liquids, just liquid seems to be a possibility. Pure solid probably not because the surface melts at a temperature lower than when the bulk melts. $\endgroup$
    – Karsten
    Commented Apr 4 at 3:01
  • 2
    $\begingroup$ The position where ice, water and water vapour exits together (at equilibrium) completely defines the triple point. The freedom of the system is zero, none of them can be arbitrarily chosen, so I agree with you that D is the answer. $\endgroup$
    – porphyrin
    Commented Apr 4 at 7:31
  • $\begingroup$ In answer to your addition, I think that the first paragraph is just wrong if the T and p quoted are the triple point. The answer sort of assumes that it takes some time to reach equilibrium or there is supercooling or some such going on, i.e. 'need not condense', as if it makes a choice to do so or not! $\endgroup$
    – porphyrin
    Commented Jun 1 at 6:55

3 Answers 3


Writing an answer because it's too large to be a comment, but it is rather indirect.

I'm quite convinced the correct (or "most correct") answer is D, because it is "simple", while answers A through C have a pernicious issue that needs consideration.

First, I suspect there is a deeper problem in virtually everyone's intuitive understanding of the triple point: speaking very strictly, it is almost surely impossible to attain experimentally, and therefore it may well never have been observed. As porphyrin says, it is a zero-dimensional infinitesimal point with precisely zero degrees of freedom in a 2D state space, corresponding to identically 0% of the physically available phase space (i.e. a space of measure zero). It is born out of mathematical extrapolation, not experimental observation. No matter what trajectory your system takes in the 2D P vs. T space, you are essentially doomed to miss the point, however small your distance from it; if you ever think you hit it, you just haven't zoomed into the graph enough.

This combines with another issue: true thermodynamic equilibria in macroscopic systems take "infinite" time to reach. Of course, many times you can get 99.99999% of the way there in experimentally accessible timeframes and call that good enough, but that last 0.00001% really matters if you're talking about a space of measure zero. When someone observes a system said to be "at the triple point", what's really happened is that they got within some tiny distance of the triple point, but which technically is still in either the solid, liquid or gas phase regions; they have not allowed the system enough time to come to full equilibrium, after which in principle they would eventually display their true location in the PT phase space. Making solid, liquid and gas appear simultaneously and irrefutably in an experiment is not enough; you have to wait an enormous amount of time holding those conditions. And more relevant to the question, you could do an experiment and then say you have reached the triple point without observing all three phases simultaneously, but that would make it a kinetic observation, not a thermodynamic one (e.g. perhaps nucleation opportunities were lacking and the gas never could condense into solid or liquid, even if magically held at exactly the triple point for a finite amount of time).

In actual reality, when taking into account the whole complexity of physics and real systems, the triple point is likely not a zero-dimensional space but will instead have a higher dimensionality, because small but non-zero terms have to be added into the chemical potential equations (e.g. surface energies or what have you). Wikipedia even alludes to this indirectly, citing systems where more phase dimensions can exist. However, even in the best case, the triple point will very likely still correspond to a vanishingly small fraction of the available phase space, and being "at" the triple point may have little rigorous meaning when taking into account experimental uncertainties.

These latter considerations could in principle allow answers between A and C to be technically possible, however they almost surely go too deep and beyond the scope of the original question, and I don't see the value of taking them into account at this level of discussion.

  • 2
    $\begingroup$ There can be (and perhaps has to be) taken frequent physical/chemical "good enough" approach, due fundamental accuracy uncertainties. $\endgroup$
    – Poutnik
    Commented Apr 4 at 10:52
  • 2
    $\begingroup$ Notwithstanding your points the triple point of water at least cannot be as difficult to determine as you suggest as it was used until 2019 as part of the definition of the Kelvin temperature scale. For example, using densities and heats of fusion and vapourisation the Claypron and Claussius-Clapyron equations could be used to extrapolate to the triple point. $\endgroup$
    – porphyrin
    Commented Apr 7 at 13:07
  • $\begingroup$ @porphyrin Indeed determining the position of the triple point is not too difficult, the difficultly is purely in sitting on the point. Perhaps I didn't convey the distinction well enough. $\endgroup$ Commented Apr 7 at 21:46

The phase transition curves and triple points (there can be more than one with multiple solid phases) on phase diagrams tells us that at these conditions, two or three phases may coexist at equilibrium.

It does not say they must be all present.

  • At the s-l curve, there may be a solid or liquid phase or both.
  • At the s-g curve, there may be a solid or gaseous phase or both.
  • At the l-g curve, there may be a liquid or gaseous phase or both.
  • At the triple point, there can be any combination of the present and missing involved three phases.

But be aware that scenarios with missing phases are special cases and often hypothetical.

  • 1
    $\begingroup$ Wait, so why do gases have to be present at the triple point according to this question? I still don't really understand what makes gases so special. $\endgroup$
    – unstable
    Commented May 31 at 6:27
  • $\begingroup$ There is theoretically possible that at triple point conditions, solid, liquid or both phases fill completely the container. But I do agree that it is rather hypothetical scenario,so let say gaseous phase is negligible. An we could still say there are created and removed randomly some miniscule amount of other phases. $\endgroup$
    – Poutnik
    Commented May 31 at 7:38
  • $\begingroup$ Imagine isothermal bath at triple point T, with piston at triple point pressure. Infinitesimal increase of pressure will cause slow condensation of gaseous phase to liquid and/or solid phase until no gaseous phase remains. $\endgroup$
    – Poutnik
    Commented May 31 at 7:55

We also should be looking at the molecular aspect which classical thetmodynamic laws miss, particularly with respect to the vapor phase. In this answer the vapor pressure of gallium at its triple point is so low that according to the Ideal Gas Law, a single molecule would require a cube hundreds of meters across to exist in equilibrium. Experimental samples are in general not that large and even if they were, they might not be equilibrated over that length scale.

The problem then becomes one of determining how probable any molecule is to occur within our sample vapor space. In the case of gallium at its triple point this probability is so low that we are very unlikely to observe a vapor phase empirically.


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