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Reversible and Irreversible reactions

  1. Oct 4, 2013 #1
    There are 2 types of reactions 1) Being fully irreversible and 2) reversible

    So for 1) the graph of Gibbs energy against the progress of reaction would look like this: http://postimg.org/image/phycri0vr/

    So the difference between 100% product and 100% reactant is the ΔG of the reaction which makes sense.

    But for 2) the graph of Gibbs energy against the progress of reaction would look like this: http://postimg.org/image/grmh7oknz/

    So now the ΔG° would be for 1 mole of reactant to form 1 mole of product and also, the actual difference is just the G value of the products minus the G value of the reactants. However, during equilibrium the ΔG is now 0. However, if we just take the difference between the G value of the equilibrium point, E and the 1 mole of reactant, that value is non-zero so why is it always known for the ΔG at equilibrium to be 0?

    Lastly, I also read that no reaction is fully irreversible so in fact all reactions has the graph on 2). However, in that case if all reaction has an eventual ΔG of 0 then how can any work be done? And we always say that using the formula ΔG°=ΔH°-TΔS° would give us the maximum work being done. But since all reactions would have an actual ΔG of 0, how can there any work be done?
     
  2. jcsd
  3. Oct 5, 2013 #2
    If you have a chamber in which the reactants are in equilibrium with the products, and you reversibly add pure components in stoichiometric proportions, and reversibly remove pure products in the corresponding stoichiometric proportions, the change in free energy between the pure reactants added and the pure products removed will be zero, and the reversible amount of work that is done in this process will be zero. The pure reactants can be envisioned as being added using semi-permeable membranes, and the pure products can be removed using semi-permeable membranes.

    Chet
     
  4. Oct 6, 2013 #3
    Hi thanks for the reply Chet,

    But since all reactions have a dG of zero, then how can any useful work be done? Or when we use the dGo=dHo-TdSo formula to get the maximum work done, isn't that value also untrue?
     
  5. Oct 6, 2013 #4
    The reactants and products in the chamber are in chemical equilibrium at the given temperature and pressure. The pure reactants are added at the same molar free energy as the partial molar free energy of the reactants in the chamber, and the pure products are removed with the same molar free energy as the partial molar free energy of the products in the chamber. The free energy of the pure products minus the free energy of the pure reactants is zero, and there is no change in free energy of the gases in the chamber.

    If we are dealing with ideal gases within the chamber at chemical equilibrium, for example, the pure reactant gases can be added at the same partial pressure as the reactant gases in the chamber, and the pure product gases can be removed at the same partial pressure as the product gases in the chamber. Provided that the reactant and product gases are added and removed in stoichiometric proportions, the free energy of the products removed minus the free energy of the reactants added is zero, and no net work is done in adding and removing gases from the chamber.

    Reactions carried out at constant temperature and pressure have dG of zero only if the pure components being added and removed through the semi-permeable membranes are at a combination of equilibrium pressures. However, if the pure components all start out at the same pressure as the total chamber pressure and the pure reactants all end up at the same pressure as the total chamber pressure, then ΔG is not zero. For example, if we reversibly expand the reactant gases to their equilibrium partial pressures in the reactor before introducing them, and the product gases are irreversibly compressed from their equilibrium partial pressures to the total reaction pressure after removing them, there will be net work done, and ΔG will not be equal to zero.
     
  6. Oct 6, 2013 #5
    Hi Chet thanks for the reply.

    I didn't quite understand a lot of them actually. I think it's because my course has only taught me the basic about thermodynamics. My module only covered the formula ΔG=ΔH-TΔS and that at equilibrium, ΔG=0 and that if we use the ΔG° we can find the Keq of the equilibrium. So from the first equation we were just taught that if the ΔG is less than zero it indicates spontaneity while more than zero indicated non spontaneity. Also, we were told that no reactions actually proceed only at one direction so they all proceed to equilibrium. So I couldn't really fit in the two things together so after thinking this through again I have 2 questions.

    1) for ΔG° do we assume that all the of the reactants gets converted to products (completely irreversible) as well as that they occur under standard state. So for example if 2A→B then for ΔG°, if would be the change in Gibbs energy from 2 moles of A to 1 mole of B under standard state conditions?

    2) So I thought initially that if we just let a reaction take place, the ΔG of the reaction would just go down to 0 no matter what. So I thought if that were the case, wouldn't the first equation we learned be meaningless? Because how can it tell us the maximum work if the reaction doesn't actually proceed in an irreversible manner?

    Thanks so much for the help
     
  7. Oct 6, 2013 #6
    Not exactly that. ΔG° is the free energy of one mole of pure B at standard state conditions minus the free energy of two moles of pure A at standard state conditions. ΔG=0 is the change in free energy you get if you have a chamber in which A and B are already in equilibrium at pressure P and temperature T, and you introduce 2 moles of pure A into the chamber at the same time that you remove 1 mole of pure B from the chamber, each at their equilibrium partial pressures within the chamber (while holding the total pressure and the temperature of the chamber constant).

    If there were an initial driving force for a chemical reaction to occur, then the system would not initially be at chemical equilibrium. But, if you were willing to ignore that and assume that there were some constraint present that prevented the reaction from occurring until the constraint was released, then you could establish the initial free energy of the system. If you then allowed the reaction to proceed at constant temperature and total pressure until chemical equilibrium was established and calculated the free energy of the system in the final state, you would find that it is lower than the initial free energy.
     
  8. Oct 7, 2013 #7
    Something that might help your understanding a lot is to look up "van Hoff's equilibrium box."
     
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