Posted by Aust on Sunday, October 14, 2012 at 3:05pm.
Consider the reaction
PCl3(ℓ) → PCl3(g)
at 298 K. If ∆H◦
is 32.5 kJ/mol, ∆S
◦
is
93.3 J/K mol, and ∆G
◦
is 4.7 kJ/mol, what
would be the boiling point of PCl3 at one
atmosphere?

Chemistry  DrBob222, Sunday, October 14, 2012 at 8:51pm
dGo = dHo  TdSo
Substitute for dH and dS, set dGo = 0, and solve for T (in kelvin).
dGo = 0 because that is that point that the liquid phase/gas phase is in equilibrium and dGop = 0.
The number you obtain is based on dHo and dSo not changing very much from the value you get from the tables at 298 K.
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