Entropy and Free Energy

1. Apr 28, 2009

MonsieurWise

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. Apr 28, 2009

MonsieurWise

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

3. Apr 30, 2009

chemisttree

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.

4. May 15, 2009

MonsieurWise

Oh, it was in my chemistry book...

5. May 15, 2009

Mapes

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, $\Delta G_\mathrm{reactant}$ and $\Delta G_\mathrm{product}$ are equal and opposite.

6. May 18, 2009

chemisttree

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?

7. May 18, 2009

Mapes

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