In Gibbs Free Energy, why multiply -T?

  • #76
I haven't heard that term used with Gibbs free energy as I really don't know much about this. But I've learned that when ΔG is positive, a favorable (spontaneous) reaction will occur and non-spontaneous reactions when its negative.
opposite you said: negative favorable or spontaneous; positive unfavorable or non-spontaneous. You could use also negative possible; positive impossible. Or negative allowed; positive prohibited
 
  • #77
Sorry I meant favorable reactions at negative Gibbs Energy. Also are you saying that reactions only occur at negative Gibbs energy?
Better say negative (or positive) DELTA Gibbs energy, since we deal with changes.
 
  • #78
A positive standard Gibbs free energy change does not mean that the reaction will not occur at all. It simply means that the conversion at which the reactants will be at equilibrium with the products will be small.

Chet
Or we could say that at standard P, T, and, most important, 1 M all participants, it will not occur at all, and worst, you will see the opposite: reactants increase above 1 M and products decrease bellow 1M.
 
  • #79
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82
Or we could say that at standard P, T, and, most important, 1 M all participants, it will not occur at all, and worst, you will see the opposite: reactants increase above 1 M and products decrease bellow 1M.
Is it right?
 
  • #80
Chet's just pointing out that Gibb's free energy has nothing to do with whether a reaction "goes" or does "not go." It limits the degree of "completion" of a reaction, allowing quantitative calculation of quantities/concentrations of products.
I think that since you write (substance A) arrow (substance B) you refer tot A as reactant and B as product, so A must decrease and B increase IF your reaction is possible. DeltaG negative will say that you correctly assumed reactants and products, and your reaction is allowed to "go".
 
  • #81
You seem to be ascribing some special significance to the situation where the equilibrium constant is unity. This value of K is no more special than any other value of K. Remember also that the standard free energy change refers to the change in free energy starting out with the pure reactants in stoichiometric proportions at 1 atm., and ending up with the pure products in corresponding proportions at 1 atm. In the equilibrium constant K, you are talking about the reactants and products not necessarily in stoichiometric proportions nor at 1 atm pressure, and the partial pressures of the reactants and products even don't have to be equal; in fact, even, if there are only single moles involved, the only requirement is that the product of the reactant partial pressures must match the product of the partial pressures of the products. If the final number of moles is different from the initial number of moles, not even this is required.

Chet
And remember that say "Why system has no available energy to do work if it's reactants and products have same concentration?" is wrong if you forget that you are talking about equilibrium constant equal 1. I think there is no example of reactants and products at same concentration and the system at equilibrium.
 
  • #82
Thermodynamics only determines whether or not a reaction can occur and what the equilibrium will be at completion. The rate at which a reaction occurs is determined by kinetics. So oxygen and hydrogen don't spontaneously react due to reasons of kinetics -but the thermodynamics is correct in that it is a downhill reaction.
 
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