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.
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?
1 answer
2 answers