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Consider the reaction
PCl3(ℓ) → PCl3(g)
at 298 K. If ∆H◦
is 32.5 kJ/mol, ∆S

93.3 J/K mol, and ∆G

is 4.7 kJ/mol, what
would be the boiling point of PCl3 at one

  • Chemistry -

    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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