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Delta G of irreversible process

  1. Sep 30, 2004 #1
    Hi,
    I'm working on a problem of the thermal stability of a protein. Conventionlly, people compare protein thermal stability in terms of the Gibbs free energy difference between the native and unfolded state. So if it reversibly falls apart, then for N <==> U, DG(N-U) is accessed from the classical equilibrium constant for the process. (DG = -RT ln K)

    But, unfortunately my system unfolds irreversibly so that N --> U.

    Is the gibbs free energy defined for an irreversible process? If so, how can it be calculated without an equilibrium constant? If it's not defined, is there an equilvalent quantity which can be used?

    The best I can do so far is to look at the kinetics, which I've done and got an Arrhenius activation energy. Can I get any more information from this?

    Please Help!!
     
  2. jcsd
  3. Oct 1, 2004 #2
    irreversible

    I don't think I can help you fully with the problem.
    But there is one idea that I have about calculating state variables like Gibbs' energy for irreversible processes.

    It is not required for the system to undergo a reversible process. If your system has a definable state, then you just need to consider the initial and final states. The change in state functions will be the same whether the system undergoes a reversible or an irreversible change if the initial and final state are the same. So, you can follow any path which makes your calculations easy and leads to the same final condition. The result will be the same.

    I saw such an example in the free expansion of a gas. Even though it is not a reversible process, we could calculate the change in entropy by taking into account only the initial and final states and some other process where it leads to the same result but was more easily calculable.

    hope that helps!

    spacetime
    www.geocities.com/physics_all/index.html
     
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