# Homework Help: Decomposition of N2O5 Entropy Equilibrium

1. Apr 8, 2014

### yolo123

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.

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2. Apr 8, 2014

### Staff: Mentor

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

3. Apr 8, 2014

### yolo123

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
4. Apr 8, 2014

### yolo123

I will post another problem on this thread. I do not want to overcrowd the forum with my "pollution."

5. Apr 8, 2014

### yolo123

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

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6. Apr 9, 2014

### Staff: Mentor

Your approach is correct. I haven't checked your arithmetic, or your use of units.

Chet