Solving Equilibrium Reactions with a Given KC and Initial Concentrations

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Discussion Overview

The discussion revolves around solving equilibrium reactions using given equilibrium constants (Kc and Kp) and initial concentrations. Participants are attempting to determine the temperature at which a specific equilibrium condition is met for the decomposition of NOBr gas.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a homework problem involving the decomposition of NOBr gas and seeks guidance on how to calculate the temperature based on given Kc and equilibrium pressure.
  • Another participant suggests calculating the number of moles of gases present, indicating that the percentage decomposition and Kp should provide sufficient information.
  • A participant expresses uncertainty about using Kc to solve for concentrations and questions the validity of their approach.
  • Some participants discuss the initial and equilibrium pressures of NOBr and the derived Kp, with one participant noting a potential error in their calculations.
  • There is a suggestion to use an ICE table to organize the calculations, but discrepancies in values lead to confusion among participants.
  • One participant recalculates using hints from their professor and arrives at a significantly high temperature, prompting further questioning about the correctness of their approach and values.

Areas of Agreement / Disagreement

Participants express differing views on the accuracy of calculated values and methods used. There is no consensus on the correctness of the calculations or the approach to solving the problem.

Contextual Notes

Some participants note that the values derived from calculations appear inconsistent, particularly regarding the relationship between Kp and Kc, and the assumptions made in the ICE table setup. There are unresolved questions about the accuracy of initial conditions and derived pressures.

Lancelot59
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Homework Statement


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?
 
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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
 
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.
 
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.
 
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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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.
 
What 'other parts'?

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Oh, there was an A and B part to that question. This final one I'm stuck on is Part C
 
The initial pressure of NOBr was .214 atm, the Kp I found was 9.66x10-6.
 
  • #10
Was there any other data given in these 'other' parts?

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  • #11
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.
 
  • #13
That formula is on my formula sheet...ah well. So if \Deltan 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
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
So what went wrong? I see nothing else that failed. My calculations for the Kp and initial pressure look right.

Can you check my answers please?
 
  • #16
Pressure looks OK, but Kp seems way too small to me.

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methods
 
  • #17
Ugh...lemme re-run it.

I ended up with 9.65x10-6 again...
 
  • #18
Show.

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  • #19
Kp = [Br2][NO]2 / [NOBr]2

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

Kp = 9.66 x10-6
 
  • #20
No, where did you get these values from.

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methods
 
  • #21
The preceding part using an ICE table and some math.

Hopefully this attachment can get approved soon...

It's part 1a.
 

Attachments

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  • #22
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
I don't get it...
 
  • #24
Show your ICE table and how you calculated - for example - 0.007275.

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  • #25
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)\Deltan to get a temperature of 5957.399661oK... Is something wrong here?
 

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