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Gibbs Energy

  1. Jun 25, 2008 #1
    Well I read in a book that:
    [tex]\Delta[/tex]G= RTln (K) (Where K is the eqm const)
    H "K" is not necessarily dimensionless so how can we take a log of "K"?
     
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  3. Jun 26, 2008 #2

    MathematicalPhysicist

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    Do you mean Gibbs free energy?

    then it's given by:
    [tex]\tau log(Z_G)[/tex]
    after taking the derivative you can get the difference or differential form of this energy.
    Z_G is the Grand canonical partition function, and it's indeed dimensionless.
     
  4. Jun 26, 2008 #3

    Andy Resnick

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    The full Nernst-Planck equation is:

    [tex]\Delta\mu=RTln(\frac{[x_{i}]}{[x_{o}]})+ZF(\psi_{i}-\psi_{o})[/tex], where

    [tex]\Delta\mu[/tex] is the change in chemical potential for a particular species

    [tex][x_{i}][/tex] is the concentration of species 'x' on one side of a dividing surface (and [tex][x_{o}][/tex] the concentration on the other side)

    [tex]\psi_{i}[/tex] the electrical potential on one side of a dividing surface (and [tex]\psi_{o}[/tex] the potential on the other side)

    And R, T, Z, F the usual gas constant, temperature, charge per molecule and Faraday constant.

    It's worth understanding this equation- it governs diffusive processes of charged solutes in solution and leads to a remarkable (IMO) result: the membrane potential. There's various simplifications, it looks like you have uncharged solutes (Z = 0), and instead of [tex]\Delta\mu[/tex] you are using [tex]\Delta G[/tex], which also changes the [tex]\frac{[x_{i}]}{[x_{o}]}[/tex] term to the equilibrium constant. But, since it's still dimensionless, there's no problem.

    Does that help? This is a really fundamental concept- make sure you understand it.
     
  5. Jun 26, 2008 #4

    Mapes

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    It's often ignored, but the K inside the logarithm is divided by K0, which has magnitude of 1 and the same units. The quotient is therefore dimensionless and equal to the magnitude of K.

    EDIT: Whoops, Andy got there first with a more complete answer.
     
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