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Gibbs Energy and Equilibrium Constant

  1. Aug 24, 2009 #1
    The relation, ΔG = -RT ln(Keq)
    is ubiquitous.

    It says that for a reaction, the change in Gibbs energy is proportional to the logarithm of the equilibrium constant.

    But where does this come from?
    I've been reading through many books and haven't yet find any that derive or indicate the history of this equation. Does anyone know?

    Is this simply an empirical result? Is it derivable other than from statistical mechanics? I've found a couple places where this can be derived from statistical mechanical postulates, but is that the only way? Is that where this relation originally comes from? (I thought it was originally based in thermodynamics; am I wrong about that?)

    Does anyone know the history of this equation? Who first wrote it down? In what publication?

    If you can shed some light in where this relationship originally came from I would much appreciate it. Thank you. --Dino
     
  2. jcsd
  3. Aug 25, 2009 #2
    I'm pretty sure it was rigorously derived for ideal gases. Pardon my rambling while I try to remember what I can...
    dU=dq+dw=TdS-pdV
    G=U-TS+PV,
    so dG= VdP-SdT
    Since V=nRT/P, this gives dG=nRT d(ln P) - S dt
    Dividing by n to get molar quantities,

    [tex]dG=RT d(ln P)-S dT[/tex]

    For a constant-temperature process we can drop that last term.
    so for a single component, we can integrate both sides to get the change in ΔG when we expand or compress a gas isothermally:

    [tex]G_{actual}-G_{standard state}=RT ln{{P}\over{P_{standard}}}[/tex]

    Now combine that for all products and reactants in a reaction, and set standard pressure equal to 1 atm and you'll get

    [tex]\Delta G_{actual}-\Delta G_{standard state}=RT ln{q}[/tex]

    Since the actual ΔG at equilibrium is zero (otherwise it wouldn't be at equilibrium), and [tex]q=K_{eq}[/tex] at equilibrium this simplifies to

    [tex]-\Delta G_{standard state}=RT ln{K_{eq}}[/tex]

    or the more famous

    [tex]\Delta G_{standard state}=-RT ln{K_{eq}[/tex]

    Ideal solutions should behave similarly (replacing pressure with concentration, although I couldn't derive it at the moment).
     
    Last edited: Aug 25, 2009
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