Solving Equilibrium Reactions with a Given KC and Initial Concentrations

In summary: Kp should be a function of temperature and not concentration. You should also use molarity or molality in your calculations. --So the final temperature is 59.57oK. Is that what you were looking for? In summary, at equilibrium, 34% of the original gas in a flask decomposes to produce a total pressure of 0.25 atm.
  • #1
Lancelot59
646
1

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) [tex]\Leftrightarrow[/tex] 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:
Physics news on Phys.org
  • #2
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.

--
methods
 
  • #3
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
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
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.

--
 
  • #6
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
What 'other parts'?

--
 
  • #8
Oh, there was an A and B part to that question. This final one I'm stuck on is Part C
 
  • #9
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?

--
 
  • #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) [tex]\leftrightarrow[/tex] 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 [tex]\Delta[/tex]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
Can't be - 0.3K is so close to absolute zero you can be sure all substances will be liquid, not gaseous. Even helium.

--
 
  • #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.

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

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

--
 
  • #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.

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

Hopefully this attachment can get approved soon...

It's part 1a.
 

Attachments

  • scanned_0022.jpg
    scanned_0022.jpg
    36.9 KB · Views: 275
  • #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.

--
 
  • #23
I don't get it...
 
  • #24
Show your ICE table and how you calculated - for example - 0.007275.

--
 
  • #25
I re-did it this morning with some hints from my Prof:

2NOBr [tex]\leftrightarrow[/tex] 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)[tex]\Delta[/tex]n to get a temperature of 5957.399661oK... Is something wrong here?
 

What is equilibrium?

Equilibrium is a state in which the forward and reverse reactions of a chemical reaction occur at equal rates, resulting in no net change in the concentrations of reactants and products.

What is the equilibrium constant (Kc)?

The equilibrium constant is a numerical value that represents the ratio of the concentrations of products to reactants at equilibrium for a chemical reaction. It is specific to a particular reaction at a given temperature.

How do I solve for equilibrium concentrations using Kc and initial concentrations?

To solve for equilibrium concentrations, you can use the equilibrium expression (Kc) and set it equal to the ratio of products to reactants. Then, plug in the given initial concentrations and solve for the unknown equilibrium concentrations.

Can I solve for equilibrium concentrations without knowing the equilibrium constant?

No, the equilibrium constant is necessary to solve for equilibrium concentrations. It is a key parameter that describes the behavior of a chemical reaction at equilibrium.

What factors can affect the value of Kc for a reaction?

The value of Kc can be affected by changes in temperature, pressure, and the concentrations of reactants and products. It is also influenced by the stoichiometry of the reaction and the presence of catalysts or inhibitors.

Similar threads

  • Biology and Chemistry Homework Help
Replies
5
Views
1K
  • Biology and Chemistry Homework Help
Replies
3
Views
2K
  • Biology and Chemistry Homework Help
Replies
4
Views
1K
  • Biology and Chemistry Homework Help
Replies
4
Views
4K
  • Biology and Chemistry Homework Help
Replies
5
Views
4K
  • Biology and Chemistry Homework Help
Replies
4
Views
2K
  • Biology and Chemistry Homework Help
Replies
1
Views
1K
  • Biology and Chemistry Homework Help
Replies
7
Views
1K
  • Biology and Chemistry Homework Help
Replies
1
Views
1K
  • Biology and Chemistry Homework Help
Replies
1
Views
2K
Back
Top