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Entropy and equilibriuum

  1. Jan 14, 2012 #1
    Given that at equilibrium total entropy change = 0
    how does the following equation make sense
    ΔS(total) = R*lnK
    if ΔS(total) is 0???
  2. jcsd
  3. Jan 14, 2012 #2


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    Staff: Mentor

    Units don't match, so it doesn't make sense. Units of entropy are JK-1, units of ideal gas constant are JK-1mol-1.
  4. Jan 14, 2012 #3
    no units of entropy are J K-1 mol-1
  5. Jan 14, 2012 #4
    oh Standard entropy = R lnK
    my bad

    nonetheless how can a standard entropy change of zero have an equation R*lnK
  6. Jan 14, 2012 #5
    Questions to ask yourself:

    1.) What is K in your equation?

    2.) What does K equal when the system is at equilibrium?

    If you do this, the answer will fall right into your lap.
  7. Jan 14, 2012 #6
    1) K is the equilibrium constant
    2) when the system is at equilibrium K = 1

    ...so your saying that ONLY at the point of equilibrium does total entropy change = zero
    ....i am still confused
    sorry :(
  8. Jan 14, 2012 #7
    in fact K does NOT nessecaryily = 1 at equilibrium
    ...therefore how can the two be related
  9. Jan 14, 2012 #8
  10. Jan 14, 2012 #9
    I have to say I've never seen this particular twist in presentation of thermo before, which is why my earlier "shot-from-the-hip" answer isn't right. And I should probably read more carefully....

    It's as if your text/reference material is trying to rework what is normally seen via Gibbs free energy statements in terms of entropy for whatever inexplicable reason - if you'll remember, we have the well-known equality

    ΔG = ΔG° + RT*ln(Q)

    where ΔG = 0 at equilibrium, and as Q → K at equilibrium, we have

    ΔG° = -RT*ln(K).

    It seems as if your text is trying to do something similar in terms of ΔS and ΔS° for whatever reason, such that

    ΔS = ΔS° - R*ln(K),

    so when ΔS = 0,

    ΔS° = R*ln(K).

    I suppose it's valid, although I've never seen it presented this particular way, at least that I can recall.
  11. Jan 15, 2012 #10
    what is the difference between ΔS° and ΔS
    and are they both calculated in the same way
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