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2 questions in equilibria

  1. Apr 8, 2010 #1
    Hi, there have been two questions that have been bothering me and I would really like some explanations.

    1. When an inert gas is added to an equilibrium reaction of gases. Why is there no change in equilibrium position? If an inert gas is added it doesn't react but increases total pressure of gases and because partial pressure is = number of moles of gas/total moles X total pressure, wouldn't the partial pressure be affected and therefore equilibrium position?

    2. In an equilibrium reaction where the G standard of the products is higher than the G standard of the reaction, there would be an increase of free energy which is nonspontaneous. Say an equilibrium reaction like this occurs initially with only just the reactants, why would the reaction proceed in the first place? What drives the forward reaction when it is not spontaneous?

  2. jcsd
  3. Apr 9, 2010 #2


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    Staff: Mentor

    No, partial pressure doesn't change. You increase both total pressure AND number of moles, these things cancel out and partial pressure stays the same. See 6th post here.

    Can you find an example of such reaction?

  4. Apr 9, 2010 #3

    Can you find an example of such reaction?


    dG standard for N2O4=97.8kJ
    dG standard for NO2=51.3kJ

    dG rxn= 51.3X2-97.8=+4.78kJ

    Why would this reaction proceed in the first place?
  5. Apr 19, 2010 #4
    The Gibbs energy change of your reaction is low so your reaction is reversible (not thermodynamically reversible), the sign of [tex]\Delta G_\textrm{rxn}[/tex] and its magnitude are important to determine the equilibrium state of a given reaction. The second law of thermodynamics establishes that this reaction, at T and P fixed, will tend to minimize the reaction Gibbs energy change as a function of the reaction's extent. I hope this helps...
  6. Apr 19, 2010 #5
    For some reactions, the gibbs free energy looks like this:


    This is the case whether gibbs free energy increases or not in the reaction. It will go to the stage of the reaction at which the gibbs free energy is at a minimum. At this point, ΔG=0.

    For reactions with activation energies, the gibbs curves look like this:


    In this case it will not proceed regardless if it is spontaneous or not, until sufficient energy is provided by the environment.
  7. Apr 20, 2010 #6
    Sorry I do not understand the difference between thermodynamically reversible and reversible. My chem standard is only up to 1st year university. Can you explain this???

    Thank you for your graphs. What I do not get though is why there is a minimum curve. I understand that in the forward reaction there causes a curve due to the increase in entropy of the mixing of the gases. But I do not understand why this is the case for the backwards reaction. Can you explain further please?
  8. Apr 21, 2010 #7
    Simply put, a container filled with either pure product or pure reactant has a higher potential energy than a container containing a mix of the two in some molar proportion. Where the point of lowest potential energy is on the reaction curve is given as a function of the concentrations of the compounds which influence the reaction. That is, by the equilibrium constant, which every reaction has.

    Here is an example. You are more than likely familiar with the fact that is you put silver nitrate into a chloride solution, you get out solid AgCl. Still, when you put solid AgCl in a body of destilled water, you can be sure that some os the salt will go into solution. The reason is that the AgCl salt in a (very) diluted AgCl solution is energetically favourable to AgCl salt in deionized water.
  9. May 14, 2010 #8
    I used the adjective thermodynamically reversible to refer to those processes in which reversible work may be applied. And a reversible reaction may be brought to its initial conditions like a rechargeable battery. Please, see

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