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Homework Help: Therhelp modynamics and the equilibrium constant

  1. Apr 26, 2006 #1
    Therhelp !!!! modynamics and the equilibrium constant

    Thermodynamics and the equilibrium constant

    Okay I have a question about the equation (* means change) (*() means stander condition) so [G=*(G)+RT ln (Keq)] or just equation *(G)= - RT ln(Keq)
    So what dose the (Keq) stander for??? Kc? Kp? Ksp?? Kf or Kd

    Is it always Kc regardless the chemical reaction? (that’s what my prof told me)
    And then use the equation Kp=Kc (RT)^*n to convert K value???

    Or is does Keq depends on the chemical reaction Kc for solution Kp for Gas???(from the text book)

    And can someone please derived the equation [G=*(G)+RT ln (Keq)] for me?
    Where is it come from and why does it makes sense?
  2. jcsd
  3. Apr 27, 2006 #2


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    The derivation is a bit extensive, and you'll have to cover some ground in Physical Chemistry before you can make sense out of this particular topic.

    The Kc,Kp,Ksp,Kf,Kd are the equilibrium constants with respect to the concentration Kc, and pressure Kp; the rest are specific forms uniquely suited to the reaction dynamics. That is, they be in terms of the concentration or pressure. Ksp is in reference to the solubility, Kf to complex ion formation, Kd to dissociation.
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