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Decomposition of N2O5 Entropy Equilibrium

  1. Apr 8, 2014 #1
    Please see attached picture.
    I need verification of my answers. I unfortunately found these problems on an old book with no answer. I would really appreciate it.

    (a) Ok. For this one, I am really not sure. PLEASE help.
    I get a very complicated formula.

    (1+x)(x^4)/(1-x)^2=Kp.

    Now, this makes no sense because I am not able to solve more than quadratics. This book should represent problems I could see on exams where they do not expect me to do more than quadratics. What do I do?

    (b) DG=DGnaught+RTln(Q)
    Q=Kp so DG=0.
    DGnaught=28058J
    DGnaught=DH-TDS.
    DS=258mol/K.

    (c) Reaction is endothermic. Adding temperature will make reaction go forward more.
     

    Attached Files:

  2. jcsd
  3. Apr 8, 2014 #2
    Part (a) does not look like it was set up correctly. If x is the final pressure of the NO2, the final pressure of the O2 is 1+x/4, and the final pressure of the final pressure of the N2O5 is 1-x/2. I guess you have to assume that the container is operating at constant volume.

    Assume that x is small compared to 1, and see what you get. If you feel that you need to get better accuracy, solve by successive substitutions.

    Chet
     
  4. Apr 8, 2014 #3
    Chet, I corrected the situation:
    I get a final x of 0.0147M. Please check my math.
    (1+x)((4x)^4)/(1-2x)^2=Kp

    Thank you so much.
     
    Last edited: Apr 8, 2014
  5. Apr 8, 2014 #4
    I will post another problem on this thread. I do not want to overcrowd the forum with my "pollution."
     
  6. Apr 8, 2014 #5
    Here is my attempt: calculate DGnaught using values (final - initial).
    Calculate Kp noting that DG=0. So -DGnaught=RTln(Kp).
    Use Clausius Clapeyron law to find new Kp at other temperature.

    Kp 298K: 5.29x10^34
    Kp 345K: 8.61x10^29
     

    Attached Files:

  7. Apr 9, 2014 #6
    Your approach is correct. I haven't checked your arithmetic, or your use of units.

    Chet
     
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