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. $\endgroup$ 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
    Apr 4 at 3:01
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    $\begingroup$ 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
    Apr 4 at 7:31
  • $\begingroup$ I can hypothetically imagine a system at the triple point or at a phase transition curve where less than 3 or 2 phases are present, if just have approached the triple point or the curve from the respective areas or curves of the diagram. // I understand the phase change curves and the triple points are states where 2 or 3 phases can coexist in the equilibrium, not as they much be present. Like a moment an additional phase(s) are already about to be created, but not yet. By other way, the zero molar fractions of respective phases are not excluded from possibilities. $\endgroup$
    – Poutnik
    Apr 4 at 9:09
  • $\begingroup$ Or the opposite hypothetical situation, when there is isothermally and/or isobarically provided/absorbed heat (covering both TP / curves cases) until one of phases just disappears. // similar scenario is a chemical equilibrium with condensed and gaseous compounds, with limiting case the condensed or gaseous phase volumes converging to zero. $\endgroup$
    – Poutnik
    Apr 4 at 9:35

1 Answer 1


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.

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    $\begingroup$ There can be (and perhaps has to be) taken frequent physical/chemical "good enough" approach, due fundamental accuracy uncertainties. $\endgroup$
    – Poutnik
    Apr 4 at 10:52
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    $\begingroup$ 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
    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. $\endgroup$ Apr 7 at 21:46

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