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

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.