Pressure and Equilibrium of Decomposition

In summary: What's the third?The third gas is O2. The mole ratio for SO2 to SO3 is 0.152. The mole ratio for O2 to SO3 is 0.180. When the volume is halved, the pressure is also halved.
  • #1
Moose100
116
0

Homework Statement


SO3 decomposes on heating:
2SO2(g)
equilibrium.png
2SO2(g) + O2(g)
A sample of pure SO3 was heated to a high temperature T. At equilibrium, the mole ratio of SO2 to
SO3 was 0.152 and the total pressure was 2.73 atm. If the piston was pushed into cut the volume in half what is the new pressure?
[/B]

Homework Equations


I tried using the ratio to get the pressures on the left. 0.414 and 2.32 respectively.Also tried multiplying 2.73 by 2 but don't think this is the way. [/B]

The Attempt at a Solution



See above. I got to "an endpoint" with the problem but feel like it's wrong want to learn a general approach. Sorry these are my weakspot![/B]
 
Physics news on Phys.org
  • #2
Can you write an equilibrium constant for the reaction?
 
  • #3
Bystander said:
Can you write an equilibrium constant for the reaction?
With or without the numbers I found that I was using that ratio they gave me wrong.
 
  • #4
You didn't found that, you were told that ;)
 
  • #5
Which means it was found out. ;) Like some of this makes sense some of it doesn't. It's not a typical ICE problem nor do I think it really applies in this instance? But I am lost so..
 
  • #6
Moose100 said:
ICE
"ICE" is what? Intro. Chem. Eng.?
 
  • #7
Initial CHange equilibrium.
 
  • #8
"ICE? RICE?" Oh, brother --- there's no way to break this to you gently --- there is one thread in the forum on "ICE," --- and, the method seems to be something of a joke/farce/disaster --- looking at "ICE/RICE" in Wiki, I'm inclined to agree. https://www.physicsforums.com/threads/rice-tables.170018/#post-1329914

Have you had any conventional chemistry training/instruction on reactions and equilibrium constants? What they are and what they can do for you? I'd really rather not get tangled up in an edu-babble religious ritual trying to make a failed instructional approach make sense to you if we can find some way around it.
 
  • #9
Bystander said:
"ICE? RICE?" Oh, brother --- there's no way to break this to you gently --- there is one thread in the forum on "ICE," --- and, the method seems to be something of a joke/farce/disaster --- looking at "ICE/RICE" in Wiki, I'm inclined to agree. https://www.physicsforums.com/threads/rice-tables.170018/#post-1329914

Have you had any conventional chemistry training/instruction on reactions and equilibrium constants? What they are and what they can do for you? I'd really rather not get tangled up in an edu-babble religious ritual trying to make a failed instructional approach make sense to you if we can find some way around it.
Lol yes I have I am actually studying for the MCAT and am doing problems. Have a bachelor in mS in bio. LOOONG time since I took chem and analytical.

I felt that ICE wasnt useful.

To come clean. I looked at the solution and had issues with how to arrive at the molarities. I know I- is 0.1M I also "know" that I2 is 0.0013M but why use this at all? Is there a reason we use the saturation molarity in water only AND that of it in KI? this loses me.

Im pretty sound on the constant.
 
  • #10
Have threads got mixed up? I know there was another one about iodine precipitating. #1 is about sulphur oxides and oxygen.

You ask about a general approach. Do you know/remember the law of chemical equilibrium, what an equilibrium constant is? Even if you don't, look it up and find It's nothing very difficult. Then what is the relation relating concentrations of these three substances in this case? Already asked in #2!

Then mass conservation laws mean that however they vary in the closed vessel when you vary the pressure or volume, neither the number of sulphur atoms, nor the number of oxygen atoms varies. Here you start from SO3 (g) (please correct your formula) and the result is that once you know the concentration, we'll call it C0 of your starting SO3, and the concentration, call it x, of any of the gases at equilibrium, you easily know concentrations of all three, in terms of x and C0. (It will be a bit simpler if you let x be [O2].)

So rewrite your equilibrium equation in terms of x and C0. Solve it. Then when you halve the volume what do you know that changes how?
 
  • #11
Equilibrium conc of:

SO3: 2.73-2x
SO2: 2x
O2 : x

2x/2.73-2x=0.152

x=0.180

Eventually I get to the point where K= 0.00415

Am I in the ballpark here?
 
  • #12
Start with the reaction: 2SO3 = 2SO2 + O2.

K = [SO2]2[O2]/[SO3]2

If I give you that much, can you get where you need to go?
 
  • #13
Excuse me but i got that already.
 
  • #14
Moose100 said:
got that already.
Good.
Moose100 said:
2x/2.73-2x=0.152
You have three gases contributing to the total pressure. You have a mole ratio for two of them, and a stoichiometric ratio for a different pair.
 
  • #15
Not enough working shown. You needed to write explicitly the equation of #12, and then express it In terms of different quantities as I suggested - explicitly. I think your expression lacks the necessary brackets and in the end there should be x2 terms in it, even x3, and where is C0?
 
Last edited:
  • #16
That's the value of the ratio given in the problem.

2.73-2x is the equilibrium value (or so I hope) for SO3. 2x is that of SO2. I set this ratio SO2: SO3 equal to 0.152 to get x and subsequently the new equilibrium values. Then I use these to get k by using the equilibrium expression.

Hopefully this is OK and makes more sense?
 
  • #17
Moose100 said:
2.73-2x is the equilibrium value (or so I hope) for SO3. 2x is that of SO2.

Lasciate ogni speranza.

2.73 is a sum of three pressures - that of SO3, that of SO2 and that of O2. If 2x is SO2, and 2.73-2x is SO3, sum of their pressures is 2.73-2x+2x=2.73 - so where is the oxygen?

No matter what question you solve, all the time you are ignoring the fact amounts of different substances is combined because of the stoichiometry of the reaction. You make the same mistake over and over.
 
  • #18
If you set out your working explicitly and used the formula quoting and using the formula given you in #12 we could help you more and you could help yourself. It is not clear if you use the square of the ratio. This squared ratio equals something invovling three parameters, C, x and K. No idea how you got any of them. You need to get C from info you are given, then x from stoichiometry, only then can you get K from the ratio given, only after that can you get a new value for new C.

(As far as I can see that will involve solving a nasty equation but all the more we need explicit clarity; we'll talk about that if we get there.)
 
Last edited:
  • #19
I know how to apply the equilibrium formula but that ratio is throwing me.
 
Last edited:
  • #20
Assume partial pressures of gases to be pSO2, pSO3 and pO2. Express the ratio and total pressure using these unknowns. What does the reaction equation tell you about the pSO2 and pO2? Are they linked somehow? Can you express this dependency with a third equation?
 
  • #21
We haven't seen you take first or second step of anything anyone has suggested to you,
 
  • #22
I'm trying it just doesn't make sense to me at the moment.
 
  • #23
Borek said:
Assume partial pressures of gases to be pSO2, pSO3 and pO2. Express the ratio and total pressure using these unknowns. What does the reaction equation tell you about the pSO2 and pO2? Are they linked somehow? Can you express this dependency with a third equation?

They are linked by their stoichiometry.

When you say the ratio you mean the 0.152? Or their stoichiometry ratios?
 
  • #24
Forget numerical values for total pressure. Borek has told you that Ptotal is the sum of three partial pressures. The original problem statement gives you one relation between the equilibrium pressure of the reactant and one of the products. The balanced equation for the reaction gives you a second relation between the equilibrium pressures of the two products. Write those relationships, each as a separate equation.
 
  • #25
Ok I was thinkkng about that.

Ptot=pO2+pSO3+pSO2
 
  • #26
Okay, that's the "zeroth" equation. Now write an equation relating pSO3 to pSO2.
 
  • #27
pSO2/pSO3=0.152 ?
 
  • #28
Good enough for the moment. Now an equation relating pSO2 to pO2.
 
  • #29
I think I see where you are going..(hopefully :wink:)

pO2=1-pSO2

pSO2=0.152pSO3
 
  • #30
Moose100 said:
I know how to apply the equilibrium formula but that ratio is throwing me.

Moose100 said:
I think I see where you are going..(hopefully :wink:)

pO2=1-pSO2
No! I thought you had at least that right before.
 
  • #31
Moose100 said:
pO2=1-pSO2
What's the stoichiometric relationship from the equation for decomposition?
 
  • #32
2:2:1
 
  • #33
Yknow what I will be back. I'm split between this and work. I will concentrate better and save you all some aggravation.:wink::-p:smile:
 
  • #34
How many moles of O2 are produced per each mole of SO2?
 
  • #35
1 mole of O2 for 2 moles of SO2
 

Similar threads

  • Biology and Chemistry Homework Help
Replies
5
Views
1K
  • Biology and Chemistry Homework Help
Replies
1
Views
2K
  • Biology and Chemistry Homework Help
Replies
10
Views
4K
  • Biology and Chemistry Homework Help
Replies
5
Views
1K
  • Biology and Chemistry Homework Help
Replies
3
Views
2K
  • Biology and Chemistry Homework Help
Replies
3
Views
3K
  • Biology and Chemistry Homework Help
Replies
1
Views
1K
  • Biology and Chemistry Homework Help
Replies
4
Views
2K
  • Biology and Chemistry Homework Help
Replies
6
Views
7K
  • Biology and Chemistry Homework Help
Replies
9
Views
2K
Back
Top