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Relation between gibbs free energy and equilibrium constant

  1. Apr 28, 2013 #1
    I am familiar with the equation ΔG=ΔG°+RT ln(Q).But I can't derive it.We have to use the equation to derive nernst equation. So please help.
  2. jcsd
  3. Apr 29, 2013 #2
    I don't remember the derivation of the above equation off the top of my head, but I'm sure you can google it or look in any intro to the Thermodynamics or Physical Chemistry text.

    As far as deriving the Nernst equation from what you have:

    ΔG = -nFE and ΔG° = -nFE°

    plug those two into ΔG = ΔG° + RTlnQ and do some simple algebraic rearranging.

    This may not be useful to you if you are in an advanced class which requires derivations of the equations that I have taken for granted. In other words this is a pseudo-derivation applicable to a freshman level Gen. Chem. course.
  4. May 2, 2013 #3


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    In this reference , post #2 PhaseShifter states: q = Keq
    Can someone show how this is ?
    Also how does R the gas constant , Avogadro's number x Boltzamann's constant,
    apply to ΔG for solutions ?
    Last edited: May 3, 2013
  5. May 3, 2013 #4


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    Staff: Mentor

    Do you know what Q is? (To be honest using q is IMHO confusing).
  6. May 4, 2013 #5


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    Thanks, was looking at q as in heat. There is not a question on the equivalence of ΔG for
    ΔG = -RTlnKeq and ΔG = ΔH - TΔS as Ill show for anyone who wants to see it
    N2 + 3H2 --> 2NH3
    Keq = (NH3)2/(N2)(H2)3 = 6.73 * 105 , lnK=13.4 , RT= 2.473 kJ/m So ΔG = -RTlnK = -33kJ/m
    In agreement with ΔG = ΔH - TΔS
    ΔH = -92kJ/m, T = 298K , ΔS = -198J/m
    -33kJ/m = -92kJ/m + 59kJ/m
    Last edited: May 4, 2013
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