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Entropy and Free Energy

  1. Apr 28, 2009 #1
    I'm having trouble understanding why in a reaction, when the free energy G of the product equal the free energy G of the reactant, the reaction is at equilibrium. Here, as my book say, the system has reached its minimum free energy. I don't really get why... Could someone explain to me...? Thanks!!
     
  2. jcsd
  3. Apr 28, 2009 #2
    Oh...never mind. It seems like the reaction just like the Delta G, not G, to be negative to be spontaneous, right? Now The only thing I don't get is:
    Since:
    Gtotal = Greactant + Gproduct
    when Greactant decrease, Gproduct increase, so how can it say at equilibrium, "the system has reached minimum free energy"?
    Thanks
     
  4. Apr 30, 2009 #3

    chemisttree

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    Is it correct to say "Gtotal = Greactant + Gproduct"?

    Gibbs Free Energy is the maximum amount of useful work you can obtain from a process not an inherent energy present in either the products or reactants. Where have you seen this equation? It's news to me.
     
  5. May 15, 2009 #4
    Oh, it was in my chemistry book...
     
  6. May 15, 2009 #5

    Mapes

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    The Gibbs free energy G is just the enthalpy minus TS. It's definitely an inherent property (a state function) of a substance.

    MonsieurWise, the question is by what magnitude the Gibbs free energies of the reactants and the products increase and decrease. The magnitudes aren't constant; they depend on the amounts that already exist. At equilibrium, [itex]\Delta G_\mathrm{reactant}[/itex] and [itex]\Delta G_\mathrm{product}[/itex] are equal and opposite.
     
  7. May 18, 2009 #6

    chemisttree

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    And just how is that calculated? Say I've got 10 grams of hydrogen at STP... What Gibbs free energy is appropriate for that substance?
     
  8. May 18, 2009 #7

    Mapes

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    [tex](10\,\mathrm{g})\left(\frac{1}{2.016} \frac{\mathrm{mol}}{\mathrm{g}}\right)\left[0\,\frac{\mathrm{J}}{\mathrm{mol}}-\left(298\,\mathrm{K}\right)\left(131\,\frac{\mathrm{J}}{\mathrm{mol}\cdot\mathrm{K}}\right)\right]\approx -190\,\mathrm{kJ}[/tex]
     
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