I am doing physical chemistry tasks in Professor Peter Atkins book. The phase diagram of UF4-ZrF4 system that I found is in the following picture

enter image description here


  1. Where is the point that the composition x(ZrF4) in liquid is equal to 0.40 (at 900oC)?

Thought: If the point was at the intersect of the x (ZrF4) = 0,4 line with 900oC line. Then I think that the solution is merely liquid (no solid presents). However, it contradicts with the statement "At 900oC, the liquid solution ... is in equilibrium with a solid solution ..." which indicates that there is solid state of ZrF4

  1. When we are observing the solution below the (L+S) area (x(ZrF4) = 0.40) but above the tie line passing through eutectic point, is it just UF4 crystallizes or both ZrF4 and UF4 crystallizes at the same time?

  2. At 850°C, when the fraction of ZrF4 (x-axis) is equal 0,3 and 0,4. The composition of liquid phase and solid phase remains the same when drawing the tie line. How could this be explained?

  • $\begingroup$ It is not a "series" of solid solutions, it is a solid solution (sigh). And the 900C vs 850C points given are in very different parts of the phase diagram, leading to serious confusion , making the 850C discussion irrelevant to the question (sigh). What a terrible question (and I say that lovingly as a phase diagram nerd). $\endgroup$
    – Jon Custer
    Commented Jan 3 at 20:28

1 Answer 1


Ad thought:

It is not in contradiction. Notice very different compositions of both phases (0.28, 0.14), compared to $x_\ce{ZrF4} = 0.4$ at $\pu{900 ^{\circ}C}$. The latter will be all liquid until cooled down to $\approx \pu{865 ^{\circ}C}$, when the solid starts forming with $x_\ce{ZrF4} \approx 0.22$. With progressing cooling, the liquid phase gets progressively enriched by $\ce{ZrF4}$, as the solid is enriched by $\ce{UF4}$.

Ad Q2:

Notice the note continuous series of solid solutions.

In a way, the diagram is equivalent to a phase diagram x-T of two miscible liquids with azeotrope with the minimum boiling point. So the solid is the solid solution containing both.

If the liquid mixture $x_\ce{ZrF4} = 0.4$ at $\pu{900 ^{\circ}C}$ is being cooled down, then:

  • At $\approx \pu{865 ^{\circ}C}$, the solid solution containing both fluorides starts forming with $x_\ce{ZrF4} \approx 0.22$.
  • This is enriching the solid phase by $\ce{UF4}$ while the liquid phase gets enriched by $\ce{ZrF4}$.
  • When the system is further being cooled, the composition of the liquid phase and the momentarily solidified phase follows the isotherm of the current system temperature.
  • For the given temperature, these both compositions gradually increase, as lower isotherms cross both composition lines at higher and higher fraction for $\ce{ZrF4}$.
  • The system converges to its eutectic, where composition of both phases is equal, with $x_\ce{ZrF4} \approx 0.77$ and the melting point $\approx \pu{765 ^{\circ}C}$.
  • At this composition and temperature, the last remnants of the liquid phase finally solidifies.

Q3: Composition of both solid and liquid phases remain the same, comparing both states, but the ratio of both phases differ. The state with the overall higher fraction of $\ce{ZrF4}$ will have higher ratio of liquid:solid phase.

  • $\begingroup$ Just to clarify the phrase "solid solution". Should it mean 2 or more solids in a mixture? Or just miscible solids like alloy? $\endgroup$
    – Shira
    Commented Jan 3 at 10:46
  • 2
    $\begingroup$ You do know what solid means and what solution means. then you know what solid solution means = a homogenous solid phase containing 2 or more components. Note that not all alloys are homogenous, having domains of microscopic sizes of different components. $\endgroup$
    – Poutnik
    Commented Jan 3 at 10:49
  • $\begingroup$ Furthermore, when we cool down more from 865°C, the tie line indicates more enriched ZrF4 solid and ZrF4 liquid at the same time as the %ZrF4 increases to the right of the diagram. Am I making mistakes here? $\endgroup$
    – Shira
    Commented Jan 3 at 11:06
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – Poutnik
    Commented Jan 3 at 11:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.