Chemistry

Phase Diagrams and Lever Rule

Justin Hsu

At first glance, there seems to be a lot going on in a two-phase diagram. There are temperatures, percentages, different elements, and symbols you’re not used to seeing. But once you understand what you’re looking for, everything makes sense.

Here is an example of a phase diagram for an SiO₂ and Al₂O₃ mixture. On the x axis, we have the mole % of Al₂O₃ in the mixture. As you go from left to right, you are increasing the amount of Al₂O₃ in the mixture, and the corresponding percentage is the mole fraction (number of moles of Al₂O₃ divided by the total number of moles) of the mixture. Sometimes the x axis will be in mass % (weight of Al₂O₃ divided by the weight of the entire mixture).

On the y axis are the temperatures, which determine which phase you are currently in. Labeled on the graph in the different zones are the different phases. Regions of note are the mixture regions where multiple phases are present. The L + mullite (SS) region indicates that the mixture consists of liquid and solid mullite. SiO₂ (cristobalite) + mullite (SS) indicates that the mixture contains both solid SiO₂ and solid mullite, and so on. The lines separating regions are known as phase transition lines, or where the mixture changes phases at the given temperatures.

Now that you know how to read a two-phase diagram, let’s look at an example problem that came from one of the homeworks.

Lucky for us, the problem is asking us to use the same phase diagram we were just looking at! One thing that you might be wondering is what in the world “proeutectic” means. The proeutectic point on a phase diagram can be found at the temperature where the mixture transitions into a mixture of two solids. In this phase diagram, there are two proeutectic temperatures. One occurs at 1587°C, where everything above this temperature contains liquid. Once the mixture is cooled down to below 1587°C, it turns into a mixture of two solids. Another occurs at 1890°C for the same reasons. Proeutectic mullite, as opposed to normal mullite, is the mullite present in the mixture prior to transitioning into that solid-solid zone.

So the problem statement gives us 20 mol % Al₂O₃ as a starting point, but is asking for a weight fraction. We’ll need to convert mol percentages into weight percentages for this problem.

Here is how to convert a mol percentage to a weight percentage. Assuming a 100 mol mixture, 20 mol % Al₂O₃ would mean 80 mol % SiO₂. Plugging in the values, we get:

So 20 mol % Al₂O₃ is equivalent to 0.298 or 29.8 weight % Al₂O₃ in this case

Next comes the lever rule, which is a handy way of finding out the phase compositions of your elements. Starting from our 20 mol % Al₂O₃, we can trace up to a given temperature. Since we’re concerned with the proeutectic region, we’ll look at one temperature above the proeutectic temperature and one below. Here’s our lever arm for 20 mol % Al₂O₃ at 1588°C (above the proeutectic temperature):


We’ve traced a vertical line up from 20 mol % Al₂O₃ until we intersect 1588°C. From there, we cast a horizontal line both left and right until it intersects with a phase transition line. The left line intersects the liquid line at about 5 mol % Al₂O₃ (dotted line). The right line intersects the mullite line at about 60 mol % Al₂O₃. This means that out of the all the mullite present in this mixture, 60 mol % of it is Al₂O₃ (and 40 mol % of it is SiO₂). Conversely, out of the all the liquid present, 5 mol % of it is Al₂O₃. To calculate the weight fraction of mullite at 1588°C, we first need to convert 60 mol % and 5 mol % to a weight fraction, just as before:



You now have a lever arm looking something like this:

A rule of thumb to remember with lever arms is the opposite rule. If you want the weight fraction of whatever is on the left, you’d take the length of the right half and divide it by the whole length. If you want whatever is on the right, you’d take the length of the left half and divide it by the whole thing. In this case, we want our weight fraction of mullite:

Now that we have our weight fraction of mullite in the liquid + mullite zone, we want our weight fraction of mullite in the solid-solid zone. To do this, we follow the exact same procedure as before, but this time draw our lever arm at a point below 1587°C:

Our new lever arm looks like this:

Solving for our weight fraction of solid-solid region mullite, we get:

Finally, to determine the weight fraction of mullite that is proeutectic, we take our weight fraction of mullite prior to the transition temperature (in the liquid + solid zone), and divide it by our weight fraction of mullite after the transition temperature (solid-solid zone):

Our weight fraction of mullite that is proeutectic is 0.818, or 81.8%

Justin Hsu

Schedule a demo

This won’t be sitting through a boring slideshow - we like to ask questions, learn about your workflows and pain points, and get creative with proposing meaningful solutions.