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Gibbs free energy

Riverdance

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    So you know the Gibbs free energy diagram that is a parabola with delta G=0 at the lowest point? It's confusing me the more I look at it. On the left hand side, Q<Keq and the reaction is spontaneous so I understand how the reaction would proceed down to G=0. But on the R hand side, Q>Keq and LaChatelier's says it will still go to equilibrium at G=0. But how can it do this when the reaction is not spontaneous???? Please help, I really want a good understand of this. Thank you
     

    Singhp03

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      spontaneity is an extensive property(changes with conditions) and thus when reactions progress in opposite direction, in terms of Q>K, it does not necessarily mean they are not spontaneous. More over, you should think of Lachateliers principle as an analogy for a scale which has a tendency to be at equilibrium, once you add too much to one side(say product) it will try to make an adjustment(product to reactant) 'in order to achieve equilibrium.

      hope that helped
       

      shaboobly

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        you need to think of equilibrium in terms of something that is continuously happening. the reaction is proceeding in both directions, even at equilibrium; it just so happens the rate of the forward reaction is that of the reverse. at this point the energy on both sides of the reaction are balanced and deltaG is 0.

        when you add something to a solution that is already in equilibrium, according to lechatliers principle the reaction with have a have to balance out by creating more products on the other side of the reaction, whether you add it on the left or the right. when you remove a reactant, it means you are removing a product from one side of the equilibrium and it will have to adjust to create more of the removed component.
         

        Riverdance

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          Thank you for both replies. I understand lachatelier's principle but I just can't make it fit with the Gibbs free energy. If the forward reaction is spontaneous, then the reverse is not spontaneous so how can it go down to equilibrium if Q>1? If we start out with 100% products of a reaction that is forwardly spontaneous, will that go? Can you elaborate on the extensive property of spontaneity?
           
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          shaboobly

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          May 12, 2012
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            for a reaction that is at equilibrium, it is not considered spontaneous because deltaG=0 and the rates of the forward and reverse reaction are running at the same speed, so to say. when you start with 100% of one product, it will only be spontaneous if the deltaG of that reaction is negative, and will reach equilibrium when deltaG=0.

            the constants Q and K are just ratios of the amount of products/reactants. when you have a Q which is greater than 1 it means there are more products than reactants, not necessarily meaning that the reaction is spontaneous or not. the spontaneity of a reaction depends on the gibbs equation and the other energies associated with the reaction: temperature, entropy, and enthalpy.

            Q is only used when something is added or taken away from a reaction already at equilibrium. when Q>K this means there are more products than there were at the true equilibrium and this the reaction will shift to make Q=K, or in the reverse direction. when Q<K it means there are more reactants than products and the equilibrium will shift to increase Q to make it = to K and run the forward reaction to equilibrium.

            le chatlier's principle does not depend on the energies of any of the reactions, it just says that it will counter the effect on the system to make it reach equilibrium again.
             

            indianjatt

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              Spontaneous usually refers to how favorable the forward reaction is. If Q> Keq, then the forward reaction is not favorable (thus not spontaneous), but the reverse is. Since everything wants to tend towards equilibrium, a high Q will shift the reaction to the left towards EQ.

              Maybe this equation will help: Delta G = RTln(Q/keq), If Q>Keq, Delta G is positive and not spontaneous. But you could very well consider the reverse reaction and see if the reaction is "spontaneous" in that direction. You will find out that it IS spontaneous in the reverse direction(Delta G = RTln(Keq/Q) for the reverse), but as the definition of spontaneous is defined, which is how favorable the forward reaction is, it will always be said to be non-spontaneous for Q>Keq. (If that makes any sense.
               

              LoLCareerGoals

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                If you are plotting dG vs Q it is not a "parabola". Where do you see Q^2 term for it to be one anyway?

                DeltaG = deltaGnot +RT*lnQ. When deltaG = 0, K=Q, then
                deltaGnot = -RTlnK
                substituting:

                DeltaG = RT*ln(Q/K). That is called a natural log graph.

                Also spontaneity is directional. For Q>K, deltaG>0 and is spontaneous in the reverse direction (which drives Q down to K).
                 

                shaboobly

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                  If you are plotting dG vs Q it is not a "parabola". Where do you see Q^2 term for it to be one anyway?

                  DeltaG = deltaGnot +RT*lnQ. When deltaG = 0, K=Q, then
                  deltaGnot = -RTlnK
                  substituting:

                  DeltaG = RT*ln(Q/K). That is called a natural log graph.

                  Also spontaneity is directional. For Q>K, deltaG>0 and is spontaneous in the reverse direction (which drives Q down to K).

                  ALWAYS know if deltaG is positive or negative, if you can. this is the deciding factor on spontaneity, not the value of Q (which depends only on the concentrations of products and reactants) and K which never changes for a reaction, unless temperature is changed.

                  deltaG ALWAYS has to be negative to be spontaneous, if it is greater than 0, like you said, it will not be spontaneous. (nvm read it wrong, but leaving it here for other people)
                   
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