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Free Energy Change/Equilibirum

  1. Nov 30, 2012 #1
    1. The problem statement, all variables and given/known data

    In the previous question, you had to calculate the the standard Free Energy Change (Go) in order to solve for the equilibrium constant, K, for the reaction:


    N2(g) + 3H2(g) → 2NH3(g)


    This is the Free Energy measured under standard conditions, when the reaction is started with 1.0 M of each of the three gases present. Calculate the non-standard Free Energy change (G) at 298 K, given the following non-standard initial concentrations of the three gases. Answer in kJ.
    Gfo = -16.6 kJ/mol for NH3 (g) at 298 K

    initial concentration (M)
    N2 1.0
    H2 0.01
    NH3 4.5

    I use the equation for K to find the equilibrium constant of Q=20049504.95

    I then use the equation ΔG=ΔG°+RTln(Q)

    I keep getting an answer of 41643.09479/1000=41.643kJ...

    Where am I going wrong?

    ----------------------

    The spontaneity of a standard reaction, ΔGo, depends on both ΔHo and ΔSo. Given the following reaction and data table, decide if each of the statements shown below are True or False.
    Assume that ΔHo and ΔSo are independent of Temperature.

    3 O2 (g) → 2 O3 (g)

    ΔHorxn 285 kJ
    ΔSorxn -137 J/K

    My answers will be in blue

    This reaction is endothermic True
    This reaction is exothermic False
    This reaction is endergonic (ΔGo > 0) at 298 K True
    This reaction is exergonic (ΔGo < 0) at 298 K False
    This standard reaction will only be spontaneous at high temperatures (T > 2080 K) False
    This standard reaction will only be spontaneous at low temperatures (T < 2080 K) False
    This standard reaction will be spontaneous at all temperatures True
    This standard reaction will not be spontaneous at any temperature True

    I have tried this problem several times. Could anybody maybe help me and tell me which ones I might have wrong?





    2. Relevant equations

    I then use the equation ΔG=ΔG°+RTln(Q)

    3. The attempt at a solution

    my attempts are listed up above with the other parts
     
    Last edited: Nov 30, 2012
  2. jcsd
  3. Dec 2, 2012 #2
    Can anybody try to help me solve this?
     
  4. Dec 2, 2012 #3

    Ygggdrasil

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    Science Advisor

    For #1, you correctly calculate that RT ln(Q) = 41.6 kJ/mol. The problem tells you that ΔG° = -16.6 kJ/mol. Calculate ΔG.


    These answers don't seem to be consistent with each other.
     
  5. Dec 2, 2012 #4
    So I plug in the 41.6 into what equation? I'm sorry I'm just a little confused. :/

    And for the next one, should the first one be false?
     
  6. Dec 2, 2012 #5

    Ygggdrasil

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    Science Advisor

    ΔG=ΔG°+RTln(Q)

    You calculated RTln(Q) and ΔG° was given to you in the problem. All you need to do is add them to get ΔG.

    Yes.
     
  7. Dec 2, 2012 #6
    And I can find Q from using the initial concentrations given?
     
  8. Dec 2, 2012 #7

    Ygggdrasil

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    Science Advisor

  9. Dec 2, 2012 #8
    Alright. Thank you for your help (:
     
  10. Dec 2, 2012 #9
    I went and added the two together but it still says I am wrong. Could I be calculating my ΔG wrong?
     
  11. Dec 2, 2012 #10

    Ygggdrasil

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    Science Advisor

    Oh, since the reaction forms 2 moles of ammonia, ΔG° = 2 *(-16.6 kJ/mol).
     
  12. Dec 2, 2012 #11
    That worked perfect, thank you! (:
     
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