Question

Use Trouton’s rule to estimate the standard enthalpy of vaporization of liquid bromine, which boils at `59.8^@ "C"`?

Answer 1

Trouton's rule gives `Delta_(vap)barH_(nbp) = "29.07 kJ/mol"` compared to a reference value of `"29.8 kJ/mol"` on NIST.

---

As long as the substance doesn't exhibit highly-ordered interactions like hydrogen-bonding or dipole-dipole, and as long as we're between `"150 K"` and `"1000 K"`, Trouton's rule states:

`Delta_(vap)barS_(nbp) ~~ 10.5R`

where `Delta_(vap)barS_(nbp)` is the molar change in entropy of vaporization.

From this,

`Delta_(vap)barS_(nbp) ~~ 10.5("8.314472 J/mol"cdot"K")`

`~~ "87.30 J/mol"cdot"K"`

At a phase transition, the molar change in Gibbs' free energy `Delta_(vap)barG_(nbp) = 0`, so

`0 = Delta_(vap)barH_(nbp) - T_(nbp)Delta_(vap)barS_(nbp)`

Thus,

`color(blue)(Delta_(vap)barH_(nbp)) = T_(nbp)Delta_(vap)barS_(nbp)`

`~~ (59.8 + "273.15 K")cdot"87.30 J/mol"cdot"K"`

`~~ "29067 J/mol" ~~ color(blue)"29.07 kJ/mol"`

The actual value is `"29.8 kJ/mol"` from NIST.

The improved, Trouton-Hildebrand-Everett rule (the THE rule) states:

`Delta_(vap)barS_(nbp) ~~ 4.5R + Rln(T_(nbp)"/""K")`

where `"nbp"` is regarding the normal boiling point `T_(nbp)` and `R` is the universal gas constant.

From this, show that `Delta_(vap)barS_(nbp) = "85.71 J/mol"cdot"K"`, and that `Delta_(vap)barH_(nbp) = "28.54 kJ/mol"` in this case. Rather interesting though that this gives the worse approximation.