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La Chatlier & Gibbs Free Energy Contradiction?

  1. Jun 23, 2016 #1
    Say I have an exothermic reaction, whose change in Entropy is positive. (not the most common of reactions, but it can still happen)

    If I increase the temperature, by La Chatlier's principle, the reaction should move to the left.

    However, by Gibbs free energy, if I increase the temperature, the second term (dH - TdS) TdS, becomes more positive (and thus Gibb's free energy becomes more negative). So the reaction should move to the right?

    So is La Chatlier's just a guiding tool for the majority of reactions, and not always correct? Like in this situation.
     
  2. jcsd
  3. Jun 23, 2016 #2

    Charles Link

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    If you have a system that can move to a state of of higher entropy and still give off energy (exothermic), it is likely that it is not in equilibrium, and thereby Le Chatlier's principle would not apply. It's already in a reactive state. Adding heat should speed up the process.
     
  4. Jun 23, 2016 #3
    Okay so basically Gibbs free energy always applies and in this case adding heat would speed up the reaction.

    However, for a system in equilibrium increasing the temperature will shift the equilibrium constant so that the endothermic reaction occurs. This makes sense to me if you kind of think of heat as one of the reactants and products, but how is this actually derived? Is it purely experimental?
     
  5. Jun 23, 2016 #4

    Ygggdrasil

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    This fact can be derived from the equation that relates the equilibrium constant to the change in Gibbs free energy. How does increasing the temperature alter the ΔG of an endothermic reaction? How does increasing temperature alter the ΔG of an exothermic reaction?
     
  6. Jun 23, 2016 #5
    Well that depends on whether the Entropy change is positive or negative right? delta G = delta H - T deltaS

    That was the original source of my confusion
     
  7. Jun 23, 2016 #6
  8. Jun 23, 2016 #7
    If I could ask one last thing (just quoting to notify), it's about Gibbs free energy. By the equation Delta G = DeltaG(standard) + RTln(Q) where Q is the reaction quotient, DeltaG is clearly dependent on the concentration of products and reactants.

    However, the basic definition of it is DeltaG = DeltaH -TDeltaS. But isn't it true that DeltaH and DeltaS only depend the amount of product that was formed, and not the concentration of the reactants and products during formation. For example, if 1 mol of some substance was formed while the concentration of the reactants was 3x larger than if a mol of the same substance was formed with less concentration, the enthalpy change and heat released should be the same.

    So under this definition of DeltaG, how does DeltaG change depending on the concentrations if the same amount of product is formed?
     
  9. Jun 23, 2016 #8

    Charles Link

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    I am no expert on thermodynamics, but ## \Delta S ## of the reaction is ## S_{final}-S_{initial} ## and both of these quantities will depend on the concentrations of each and all of the substances that take part in the reaction=the initial concentrations and the final ones. I believe entropy is a quantity that is very much concentration dependent.
     
  10. Jun 23, 2016 #9
    In the equation ##\Delta G=\Delta G^0+RT\ln Q##, ##\Delta G## refers to the change in free energy in going from stoichiometric molar quantities of pure reactants at temperature T and prescribed pressures (not necessarily 1 bar) to stoichiometric molar quantities of pure products at temperature T and prescribed pressures (not necessarily 1 bar).
     
  11. Jun 24, 2016 #10
    Yes, I understand this. However my question is, depending on the current concentration of products and reactants, does either deltaH or deltaS change? I wouldn't think deltaH would, possibly deltaS would. Obviously at least one of them would have to change, since deltaG changes as Q does.
     
  12. Jun 24, 2016 #11
    I am very puzzled by this question. If the temperature, pressure, and current concentration (whatever the current concentration means) are constant, how can H, S, and G change? Are you talking about a liquid phase reaction or a gas phase reaction? Are you talking about a van't Hoff equilibrium box, in which pure reactants are injected and pure products are removed?
     
  13. Jun 24, 2016 #12

    Ygggdrasil

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    I believe the correct equation here should be ΔG° = ΔH° - TΔS°.
     
  14. Jun 24, 2016 #13
    Yes I am talking about a van't Hoff equilibrium box. What I'm considering are two seperate equilibrium boxes. G will remain constant within the boxes, but be different between the two boxes. Both at the same temperature and pressure, but one box has a different ratio of reactants and products to the other box.

    I know that G must be different, since the reactant quotient is different between the two boxes. But I'm wondering, it doesn't really make sense to me that DeltaH(for a unit amount of reactions) would have changed much if at all. Shouldn't some pair of molecules reacting to form a new molecule release the same amount of energy, regardless of the conditions the molecules are constrained to? (Temperature, pressure, and especially relative concentration). I'm willing to accept deltaS (for a unit amount of reaction times) might change depending on concentration, but I'm wondering why?
     
  15. Jun 24, 2016 #14
    If the contents of the two boxes are each at equilibrium, then the ratio of reactants and products in the two boxes is the same, and equal to the equilibrium constant.

    When you are talking about ##\Delta H##, ##\Delta S##, and ##\Delta G##, you are referring to the change in these functions between two thermodynamic equilibrium states. It is important in such cases that you precisely define the initial and final thermodynamic equilibrium states. So please, now define for us the two thermodynamic equilibrium states that you are referring to here (so that we can remove all ambiguity).

    State 1: ......


    State 2:...

    Chet
     
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