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Q value with species in different states

  1. Mar 19, 2014 #1
    Given the half reaction:

    2H2O(l) → O2(g) + 4H+(aq) + 4e-

    How would you set up the Q equation for this reactions?

    Would it be Q = [H+]4 only or do we assume the partial pressure of oxygen gas is 1 atm? Is there a general way to write the Q value for species in different states?
     
  2. jcsd
  3. Mar 19, 2014 #2

    Borek

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    Typically we will just express the partial pressure of the oxygen in bars and concentration in M. No problem with mixing states, as technically these are just approximations of the activity.
     
  4. Mar 19, 2014 #3
    Oh really? My class didn't delve too deeply into fugacity, but I was told when assigning partial pressure values in the Q equation, to ensure units were always in atm. I wasn't told exactly why, but why do you say to use bars? Do you mind referring me to a derivation or reasoning perhaps?

    Also, in my class, the professor completed an example where he solved for Q but omitted the gaseous species involved since it was a half-reaction in a voltaic cell. It was not at standard conditions, though. Is it reasonable to assume the partial pressure was 1atm, though?
     
  5. Mar 20, 2014 #4

    DrDu

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    What is Q? I only see people talking about Q recently in this forum? Is there a new book which decided to use Q instead of K?
     
  6. Mar 20, 2014 #5

    Borek

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    Q is the reaction quotient, K is the equilibrium constant. I found it is convenient to treat them separately, as they are not always identical.
     
  7. Mar 20, 2014 #6

    Borek

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    This is just a convention. Pressure should be in bar, since 1982. atm was an earlier standard.

    Sorry, there are some problems with IUPAC servers and I can't trace it right now to any acceptable source, have it just on disk in a private discussion with people I trust.
     
  8. Mar 20, 2014 #7

    DrDu

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    Ok, reaction quotient was new to me as a concept. I still don't see what it is good for. What I want to say is that the Nernst equation is derived for an electrochemical equilibrium situation, so Q=K.
     
  9. Mar 20, 2014 #8

    Borek

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    When you mix some slowly reacting reagents Q doesn't equal K - it gets there, but slowly. Without introducing Q it is difficult to explain how come we have a solution in which... K doesn't equal K? But you can easily say Q doesn't equal K and the situation is clear.
     
  10. Mar 20, 2014 #9
    Okay, thanks for the help. If you ever do find the source you could link me to, that would be great!
     
  11. Mar 20, 2014 #10

    DrDu

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