# Homework Help: Gas problem

1. Oct 3, 2009

### Lancelot59

1. The problem statement, all variables and given/known data
1. A flask initially contains only NOBr gas. Once heated to a temperature T, 34.0 % of the original gas decomposes via the following equation to give a total pressure of 0.25 atm at equilibrium:

2NOBr(g) $$\Leftrightarrow$$ 2NO(g) + Br2(g)

Given a KC of 3.94 x 10-4 I need to find the temperature. I've already gotten the KP and initial concentration as being 9.66x10-6 and 0.213675 atm respectively. I didn't see any equations for this in my notes...so how can one go about this business?

Last edited: Oct 3, 2009
2. Oct 3, 2009

### Staff: Mentor

You have to calculate number of moles of all gases present. Seems to me like 34% and Kp should be enough for that, just remember that pressure is directly proportional to n.

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methods

3. Oct 3, 2009

### Lancelot59

Well I can assume that it's a total of 1L for volume. I do have the partial pressures for all three gasses, so I can probably just work from those. I'll post again once I try this out.

4. Oct 3, 2009

### Lancelot59

Ok, I got nowhere.

I used PV/RT = n to try and get the mols, which failed for obvious reasons.

So I tried to use the Kc to solve for it:

3.94X10-4 = [Br2][NO]2 / [NOBr]2

Can I just throw X in for each of the concentrations and solve? If so then x is equal to the Kc...that seems wrong.

5. Oct 3, 2009

### Staff: Mentor

If there were initially n moles of NOBr, how many moles will be present after 34% reacted? How many moles of both products? Can you build Kp from these values?

Hm, now I started to wonder if you have enough information for that approach, but it is 2 am here and my thinking is a little bit blurred. CU tomorrow.

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6. Oct 4, 2009

### Lancelot59

From other parts of the question, I calculated the initial pressure of NOBr, as well as the final equlibrium pressures of all the gasses, as well as the Kp.

7. Oct 4, 2009

### Staff: Mentor

What 'other parts'?

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8. Oct 4, 2009

### Lancelot59

Oh, there was an A and B part to that question. This final one I'm stuck on is Part C

9. Oct 4, 2009

### Lancelot59

The initial pressure of NOBr was .214 atm, the Kp I found was 9.66x10-6.

10. Oct 4, 2009

### Staff: Mentor

Was there any other data given in these 'other' parts?

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11. Oct 4, 2009

### Lancelot59

1. A flask initially contains only NOBr gas. Once heated to a temperature T, 34.0 % of the original gas decomposes via the following equation to give a total pressure of 0.25 atm at equilibrium:

2NOBr(g) $$\leftrightarrow$$ 2NO(g) + Br2(g)

a) Determine the original pressure of NOBr in the flask.
b) What is the value of Kp at this temperature T?
c) If the value of Kc at this temperature T is 3.94 x 10-4, determine the temperature T.

That is the whole question.

12. Oct 4, 2009

### Staff: Mentor

Last edited by a moderator: Apr 24, 2017
13. Oct 4, 2009

### Lancelot59

That formula is on my formula sheet...ah well. So if $$\Delta$$n is 1 then the answer is a decimal...

9.66x10-6 = 3.94x10-4(0.08206 T)1

T=.298778534 oK

Is that right?

14. Oct 4, 2009

### Staff: Mentor

Can't be - 0.3K is so close to absolute zero you can be sure all substances will be liquid, not gaseous. Even helium.

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15. Oct 4, 2009

### Lancelot59

So what went wrong? I see nothing else that failed. My calculations for the Kp and initial pressure look right.

16. Oct 4, 2009

### Staff: Mentor

Pressure looks OK, but Kp seems way too small to me.

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methods

17. Oct 4, 2009

### Lancelot59

Ugh...lemme re-run it.

I ended up with 9.65x10-6 again...

18. Oct 4, 2009

### Staff: Mentor

Show.

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19. Oct 4, 2009

### Lancelot59

Kp = [Br2][NO]2 / [NOBr]2

Kp = (0.003638)(0.007275)2 / (0.14124)2

Kp = 9.66 x10-6

20. Oct 4, 2009

### Staff: Mentor

No, where did you get these values from.

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methods

21. Oct 4, 2009

### Lancelot59

The preceding part using an ICE table and some math.

Hopefully this attachment can get approved soon...

It's part 1a.

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22. Oct 4, 2009

### Staff: Mentor

Unfortunately it is 1 am here and tomorrow is Monday, so I have to go to sleep.

ICE table sounds OK, but values are wrong. From the data given in question [NO]/[NOBr] = 0.34/(1-0.34) - they should be comparable when it comes to value, you have them differ almost 100 times.

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23. Oct 4, 2009

### Lancelot59

I don't get it...

24. Oct 5, 2009

### Staff: Mentor

Show your ICE table and how you calculated - for example - 0.007275.

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25. Oct 5, 2009

### Lancelot59

I re-did it this morning with some hints from my Prof:

2NOBr $$\leftrightarrow$$ 2NO + Br2

I 2x / 0 / 0
C -2x(.34) / +2x(.34) / +.34x
E 2x(.66)/ +2x(.34) / .34x

The total pressure was .25 atm, so:

0.25 = PNOBr + PNO + PBr

0.25 = 2x(.66) + 2x(.34) + .34x

X = .02136752137

Then I just plugged it into the expression for each gasses concentration, and shoved those values into the Kp equation to get .1926125012.

That I plugged into Kp=Kc(RT)$$\Delta$$n to get a temperature of 5957.399661oK... Is something wrong here?