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Keq from Gibb's Free Energy

  1. Mar 4, 2014 #1
    Just for kicks. I tried calculating Keq from Gibb's free energy. This should be a straight forward calculation, but the answer is no where near close.

    Here is my calculations in SAGE:

    sage: R=8.314; T=298; G=237000 ; Keq=var('Keq'); f=G+R*T*log(Keq)
    sage: f.solve(Keq)
    [Keq == e^(-59250000/619393)]
    sage: float(e^(-59250000/619393))

    Note the positive sign before R, that is because when I solve for Keq sage assumes f=0.
    I assume using the -(G) should give the self dissociation constant for water. That is 10^(-14)

    Chris Maness
  2. jcsd
  3. Mar 4, 2014 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    First of all, this is probably more appropriate for the Chemistry Board.

    Your post is a model of how not to explain a problem. You have not explained what reaction you are you dealing with. You have not explained how you get your formula for f or what f is. You have not explained the theory behind what it is you are trying to do. You have not explained the mathematics either. If you want someone to help you, you have to do a much better job of presenting your question.

  4. Mar 4, 2014 #3
    Yes, I see that it is poorly formatted. I have students today, and I just through it up there expecting a quick answer. The reaction is the self ionization of water [tex] { H }_{ 2 }O(l)\Leftrightarrow { H }^{ + }(aq)+{ OH }^{ - }(aq) [/tex]

    However, in my haste I over looked something. I used the Gibbs free energy thinking that the products would be elemental. I need to use the heat of formation for Hydroxide and Hydronium and try it again.

    And yes, I do agree it should be in the chemistry topic.

    Chris Maness
  5. Mar 4, 2014 #4
    I have it now. Looked up Gibbs. Not sure why H+ Standrd Gibbs of Formation is 0. All the other ions have a value. I imagine it has to do with [tex] { 2H }^{ + }+2{ e }^{ - }\rightarrow { H }_{ 2 } [/tex] has a half cell potential of Zero.

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