How would you explain the phase diagram of sulphur?

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Answer 1 · 2016-04-08T20:39:04

How about this:

A phase diagram is a chart that shows the conditions of pressure and temperature at which distinct phases occur and coexist at equilibrium.

The lines on a phase diagram divide into regions – solid, liquid, and gas.

The phase diagram of sulfur is

chemwiki.ucdavis.edu

The diagram is complicated by the fact that sulfur can exist in two crystalline forms: rhombic and monoclinic.

Let's look first at the four areas:

The corresponding curves are:

lower left to ① — the sublimation curve of rhombic $"S"$ : $"S(rhombic)" ⇌ "S(g)"$
① to② — the sublimation curve of monoclinic $"S"$ : $"S(monoclinic") ⇌"S(g)"$
② to upper right — the vapour pressure curve of liquid $"S"$ : $"S(l)" ⇌ "S(g)"$
① to ③ — the transition curve for $"S(rhombic)" ⇌ "S(monoclinic)"$
② to ③ — the melting point curve for $"S(monoclinic) ⇌ S(l)"$
③ to top — the melting point curve for $"S(rhombic) ⇌ S(l)"$

There are three triple points:

① ( $"95.4 °C", 1 × 10^"-4"color(white)(l) "atm"$ ) — rhombic $"S"$ is in equilibrium with monoclinic $"S"$ , and both have the same vapour pressure.
② ( $"119 °C", 5× 10^-4color(white)(l) "atm"$ ) — monoclinic $"S"$ melts; this is the triple point for $"S"_"m" ⇌ "S"_"l" ⇌ "S"_"g"$ .
③ ( $"151 °C, 1288 atm"$ ) — rhombic, monoclinic, and liquid $"S"$ are at equilibrium.

The critical point — where liquid and gaseous $"S"$ have the same density — is off to the right at $"1041 °C and 203.3 atm"$ .