Pressure and Equilibrium of Decomposition

[Bystander]

pO2=1-pSO2

What's the stoichiometric relationship from the equation for decomposition?

 

[Moose100]

2:2:1

 

[Moose100]

Yknow what I will be back. I'm split between this and work. I will concentrate better and save you all some aggravation.

 

[Borek]

How many moles of O₂ are produced per each mole of SO₂?

 

[Moose100]

1 mole of O₂ for 2 moles of SO₂

 

[Borek]

How are their pressures related?

 

[Moose100]

IM not able to do much at the moment I said that earlier.

 

[Borek]

Write the equation that combines these pressures

 

[Moose100]

Ptotal=pO₂+pSO₃+pSO₂

pSO₂/pSO₃=0.152

pO₂/pSO₂= 0.5

Ptotal= 0.5pSO₂ +pSO₂/0.152 + pSO₂

Ptotal=8.08pSO₂

pSO₂=2.73/8.08=0.0338
pSO₃=0.338/0.152=2.22
pO₂= 0.169

k=[pSO₂]²[pO₂]/[pSO₃]²

k=[0.338]²[0.169]/[2.22]²
k=3.9 × 10^-3

 

[Moose100]

1. If they are correct the equilibrium values become the initial values upon the change in condition. I.e. Moving the piston in halfway.

2. V₂=V₁/2
P₁V₁=P₂V₂
P₁V₁=V₁P₂/2
2P₁=P₂

3. 2SO₃ 2.22-2x
2SO₃ 0.338+x
O₂ 0.169-x



Ok this is where I spent many hours last night. One because I seem to be stuck at factoring the 2P1 concept and also I am still shaky if these are the right pressures. When using the equilibrium constant to solve for the new pressures.

k=[pSO2]²[pO2]/[pSO3]²3.9 ×10^-3=[pSO2]²[pO2]/[pSO3]²

 

[epenguin]

First check results for consistency.
You say pO₂/pSO₂= 0.5 but it isn't in your conclusions. - only a decimal point mistake or other?

Then I would like to say that - I don't know that this is always followed, but In my book equilibrium constants have to have units.
Well for reactions like A ⇔ B, or A + B ⇔ C + D the eq, constant is dimensionless, but in your case it has the dimensions of pressure and units of atm. Somebody else might use torr etc. but it could also have been concentrations and M (as I was assuming in my sketch). You could have done the whole thing in terms of concnetrations but I think it is equivalent reasoning and only units and their numbers different - hope this will be clear to you.

If and when this is right, I have not checked everything, you have done the first two steps of #14.

The next will be step 3. I think the equation is not as horrible as I first thought, but you can begin formulating that.

 

[Moose100]

Ive done so last night. I took note of you saying that its a nasty equation that has a cubed power in it lol. Before I do that do we take the ratios and modify them due to the new conditions? there's is where I am stuck. Multiply them by two so to speak. Also you mean the ratio of O2 to SO2 doesn't work? Or are you saing the ratio is wrong all the way around?

 

[Moose100]

Ahh I see what you mean the ratio is inflated by 10.

 

[epenguin]

Yes - similarly check other ratios agree with your starting ones. For rest I have to take short break now.

 

[Moose100]

Oops got my notes out it was a typo.

 

[Moose100]

First check results for consistency.
You say pO₂/pSO₂= 0.5 but it isn't in your conclusions. - only a decimal point mistake or other?

Then I would like to say that - I don't know that this is always followed, but In my book equilibrium constants have to have units.
Well for reactions like A ⇔ B, or A + B ⇔ C + D the eq, constant is dimensionless, but in your case it has the dimensions of pressure and units of atm. Somebody else might use torr etc. but it could also have been concentrations and M (as I was assuming in my sketch). You could have done the whole thing in terms of concnetrations but I think it is equivalent reasoning and only units and their numbers different - hope this will be clear to you.

If and when this is right, I have not checked everything, you have done the first two steps of #14.

The next will be step 3. I think the equation is not as horrible as I first thought, but you can begin formulating that.

The K I got is then 3.9 × 10^-3atm

 

[Moose100]

[0.338+2x]²[0.169+x]/[2.22-2x]²=3.9×10^-3atm

At equilibrium under the new conditions P₂=2P₁

That means we multiply these equilibrium expressions in the brackets by two:

(0.676+4x)²(0.338+2x)²/(4.44-4x)²=3.9*10^-3

Eventually, we get x= -0.032atm

so

0.676+ 4(-0.032)=0.548atm
0.338+ 2(-0.032)=0.274 atm
4.44 + 4(-0.032)= 4.568 atm

= 5.39 atm

 

[Moose100]

If this is correct o this isn't a fluke thank you everyone for the help. I may come back with some questions about general approach after I've scrutinized some more.

 

[epenguin]

I don't usually check peoples' arithmetic but often notice anomalies like that.

I'll assume no more mistakes. Actually figures sound to me reasonable.

Parts 1 and 2 of the problem are things I'd say should be expected of you with your background - fairly elementary physical chemistry calculations. Part 3 is more challenging. Before we get into that - and it is too late night for me to concentrate - I think equally to be expected of you is you can translate those pressures (including the eq constant) into molarities. Especially if you are preparing for an exam, plus two other reasons.

I at least find it not simple to imagine what happens to pressures when you halve the volume. It is easy to imagine what happens to total concentration of something in all its forms. If the substances were not reacting your PV laws would hold, but if they react? Also I think your work here suffered from lack of explicitness. If you had written out things like the equilibrium equation you would have saved time and mistakes in the immediate, and also in the long run as when you revise you (as well as us) may not understand why you wrote something. It would be good to write out a couple of equations connecting molarities e.g. of [O₂] (= x) and the other things.

 

[Moose100]

That's what's troubling me I tripped up early on with the ratios. Like I had no clue on how to use them for some reason to even use them at all. I am looking to do more reading on this as well. Is the 0.152 ratio..this isn't a mole fraction ratio is it? That would be a fraction of the total so probably not?

The Ptotal formula should have been easy too. It just didn't dawn on me at all. Any advice on what to read to furnish more understanding in addition to what I am looking to read I am totally open to.

 

[Moose100]

One question if this is overkill I apologize. There is a second problem that is essentially an extension of this one. It says that at equiibrium the concentration of SO₃ is doubled instead of the piston being pushed in. The term as we found out initially is 2.22-2x. When the concentration doubles is it 4.44-2x? I feel like we should subtract 4x like we did in the previous problem when we doubled pressure, but ironically with that term I solved for x and got the new pressure for SO₃ to be 4.26atm as stated in the book. I understand that concentration and pressure can be equivalent in these problems but is this more of a qualitative distinction?

One thought I have is because even when you double the starting amount it depletes by a multiple of it's stoichiometric coefficient, while when you double the pressure at equilibrium it actually the initial plus the change i.e. the overall equilibrium pressure of SO₃.

Thanks again I am learning a lot from you all. I am not taking a class this is just an independent effort in prep for the MCAT I am essentially teaching myself old and new concepts I hevent really learned.

 

[Moose100]

First thing that comes to mind with equating moles and pressure is Ptot=nRT/V I would think T and V would be constants so Ptot=nc

ITs late more tomorrow.

 

[epenguin]

I don't think any reading will do you as much good as seeing this problem through to solution now, i.e. being able to use what you know - after that maybe what you read would mean more.

That equation of #52 is OK. However checking back I see they give you neither V nor T! V is not a problem, we can do it for 1L which is then compressed (at constant temperature) to 0.5 L. You can just keep T in your equation, indeed keep RT. However to carry through calculations, since we can do without complication I suggest we work with a particular value - I suggest unless anyone knows it is unrealistic 600°K.

Before that even, there is a an imprecision if not misconception when you write the concentration [SO₃] doubles. [SO₃] is not what doubles. If everything doubles there is practically no problem, but the compression makes the equilibrium shift. You need to write out those equations I mentioned! Again explicitness. I think at a glance your equations are mistaken, and writing these others will guide you to right ones.

(Incidentally I trust you notice that SO₃ holds together fairly well and it has to be made quite thin to get appreciable dissociation.)

 

[Moose100]

Ok I see what you mean about it doubling. That language was in reference to another problem that is a continuation of the one we have done over the last few days where it says that we inject the same equilibrium amount of SO3 into the vessel. Effectively doubling the concentration. I took this eventually to mean(through trial and error really) to mean that the 2.22-2x equilibrium term to turn into 4.44-2x. Which is different than the 4.44-4x we get when the last problem doubled the pressure. I just wanted to make sure I knew it definitvely. Again I will post more on what this problem satys explicitly later when work has calmed down

 

[Moose100]

Ptotal=pO₂+pSO₃+pSO₂

pSO₂/pSO₃=0.152

pO₂/pSO₂= 0.5

Ptotal= 0.5pSO₂ +pSO₂/0.152 + pSO₂

Ptotal=8.08pSO₂

pSO₂=2.73/8.08=0.338
pSO₃=0.338/0.152=2.22
pO₂= 0.169

k=[pSO₂]²[pO₂]/[pSO₃]²

k=[0.338]²[0.169]/[2.22]²
k=3.9 × 10^-3

Ok referring to the above info there is another problem that is an extension of this one I am putting it here as promised. I think this is appropriate instead of cluttering with another thread? Please let me know:

Problem:
Consider the equilibrium described in the previous problem. Instead of pushing in the piston, a quantity of SO₃(g) equal to that already present is injected. This addition momentarily doubles the concentration of SO₃(g):

A.) What is the new SO₃(g) pressure when equilibrium is restored?

Attempt:

2SO₃ (g) ⇔ 2SO₂(g) + O₂(g)


2SO₃(g): 4.44-2x This reflects the doubling of concentration after adding the amount present doubling it.

2SO₂(g): 0.338- 2x

O₂(g): 0.169-x
3.9 × 10^-3=[0.338-2x]²[0.169-x]/[4.44-2x]²

Eventually; x=0.091

So for 2SO₃(g)= 4.44-2(0.091)= 4.26atm


Like why not 4.44-4x like in the previous problem? Is it because the 2.22atm term is the concentration at first while the -2x term is the amount it depletes by? It depletes by a multiple of it's stoichiometric coefficient.

Is it also because the 4.44-4x reflects a doubling of the equlibrium pressure?

Just wanted to post for more clarity.

 

[epenguin]

The new problem is no different from the one you haven't solved! You already realized that a twofold compression doubles the total concentration of all the matter inside the vessel. Doubling the amount by injection does just the same amount far as concentrations are concerned, just the volumes are different.

It looks to me your new equations are inconsistent with the ones you had before.

People were getting impatient with you before because (a) what is really quite a simple question that can be answered in half-a-dozen lines is extending over nearly three pages (b) this is result of how you are approching it. You seem very reluctant to write a chemical/algebraic formula when pressed, but do a lot of rather vague wordy arguing.

I would like to see you present the problem and solution steps algebraically, never mind the numbers. Give the equation/s for mass conservation, or mass balance, relating the concentrations of the three substances, the total starting concentration of SO₃.
And any other relations between concentrations. Even if some of this or similar was done already. Write in the same place the equation of the equilibrium constant because we want all the equations we are going to use in one place, not three pages away. Try and get as far as you can with them - them reduces to an equation in one 'independent variable' -original or total concentration of SO₃ which we called C, and [O₂] which we then called x which we will try to solve for ifnwenhave an equation for it.

 

[Moose100]

I put that because you were telling me I wasn't explicit enough. That's why I posted that way.

 

[epenguin]

By explicit I meant just a few equations and steps with in few words their rationale.

 

[Moose100]

I don't know man you take some genuine questions and say I'm arguing maybe if you answered them the thread wouldn't be so long? It's not like I'm asking directly for an answer. I'm not arguing I'm discussing the problem. I'm not reluctant about anything. But I'm a little puzzled by what you're asking me. I've thanked and apologized enough. I think you take another look at my previous post.

 

[Drakkith]

I've unlocked the thread and I encourage all participants to pause and take a deep breath if necessary. Tutoring is very difficult when all you have to work with is text.