Pressure and Equilibrium of Decomposition

[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 tommorrow.

 

[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.

 

[Moose100]

Thank you l. I really want to solve this.

 

[epenguin]

I will be back later this evening. But it would not do any harm to write out those equations I requested meantime.

What is the name again of that exam you are preparing? It is here somewhere but I can't find it.

 

[Moose100]

The mcat

 

[epenguin]

OK I may not be it quite so fast as promised. I have only just got through recalculating from scratch the equilibrium constant. Congratulations anyway on getting the right answer - I got it on the third round! However I can make some suggestions for clarity of presentation which will be important in an exam and actually help calculation. Will post later, maybe tomorrow.

The reason I ask you to write about a slightly different question is that this will be useful to you. It is not enough to just solve this question because this question will never come again. More usually in equilibrium questions concentrations will be involved. So the question is if you had exactly this same reaction but were given initial [SO₃] (molarity) and an equilibrium constant K, how would you frame algebraically an equation involving just one unknown x (that is [O₂])?

 

[Moose100]

For
aA ↔ cC +dD

Ive got some kind of glitch here..geez.

K=[A]^a/[C]^c[D]^d

If there is an unknown for say A and we know the rest we substitute X(or leave it as A) and solve. Knowing the rest of the values.


[A]^a=K/[C]^c[D]^d

Solve algebraically.

Also in you most recent post. The second problem was specifically saying the conc. of one of the reactants doubled so ONLY the initial doubles and not the whole term of 2.22-2x? I wrote 4.44-2x and got a correct answer. But want learn the distinction between that and the original problem of double the PRESSURE at equilibrium.(4.44-4x for this term and the others)

I think my issues come when they manipulate certain aspects of the equilibrium reaction as distinguished and described above. I feel like there is something that I need to nail down. Sometimes trial and error does that lol.

 

[epenguin]

For
aA ↔ cC +dD

Ive got some kind of glitch here..geez.

K=[A]^a/[C]^c[D]^d

If there is an unknown for say A and we know the rest we substitute X(or leave it as A) and solve. Knowing the rest of the values.

Only a niggle but (I never remember the convention myself and if I am doing a problem myself write the equilibrium constant any which way round) but although your equation is true it seems the convention is when you write the chemical equilibrium that way round you put products in the numerator. That anyone is doing that in anything you come acros you can spot by the units quoted - unless the constant is dimensionless as in something like simple A ⇔ B or two reactants two products. (Of course then which you define as products and reactants is arbitrary - I don't actually know that there is a convention about that.).

You say 'if we know the rest'. But in the given problem we don't know the rest - we start with three unknowns, [SO₃], [SO₂] and [O₂]. You have to use other information to reduce the number of variables.

Also in you most recent post. The second problem was specifically saying the conc. of one of the reactants doubled so ONLY the initial doubles and not the whole term of 2.22-2x? I wrote 4.44-2x and got a correct answer. But want learn the distinction between that and the original problem of double the PRESSURE at equilibrium.(4.44-4x for this term and the others)

I think my issues come when they manipulate certain aspects of the equilibrium reaction as distinguished and described above. I feel like there is something that I need to nail down. Sometimes trial and error does that lol.

That doesn't seem right to me. How do you know this is the correct answer - was it given? I don't know where your 2.22 comes from (it happens to be close to [SO₃]). You can't just double everything - that would also double the equilibrium constant. Check that. It works for the dimensionless equilibrium constants I mentioned above but not for one like this.

Will try to catch up with #64 tomorrow.

 

[Moose100]

oops! yes products do go there distracted by that stupid glitch!

 

[Moose100]

Well we solved the first problem when the piston was moved in halfway doubling the pressure. THEN I added another question about the same system where they added the same amount of SO3 that was there and left the piston alone. The question then asks what is the conc at equilibrium. I mean I got the answer but I don't know if it's just a fluke.

Yes the answer is given. I am using schaums college chem.

 

[Moose100]

16.32 was what we did first now I'm talking specifically about 16.33. Just for your clarity.

 

[epenguin]

Well we solved the first problem when the piston was moved in halfway doubling the pressure. THEN I added another question about the same system where they added the same amount of SO3 that was there and left the piston alone. The question then asks what is the conc at equilibrium. I mean I got the answer but I don't know if it's just a fluke.

Yes the answer is given. I am using schaums college chem.

You claim to have solved the problem of when the piston was moved. If that is right then yes, I think it is a semi-fluke. Has Schaum got the answers? Is your anwer exactly right according to them?

I see the book tells you what I have hinted, you get a higher degree equation that can in practice only be solved by approximations. Or I would have said find it with a computer or calculator device (that does exactly that). (They say successive approximations, and I doubt your semi fluke is sufficient handle for the next approximation.)

But what is the equation that you have to solve? I have asked you that half a dozen times. Reminder: equation with only one unknown. This exam is not testing for mathematical ability to solve equations (except elementary) but ability to formulate them! I have to tell you you would have failed the exam through not taking a step towards this formulation. Look back to what I said about conservation equations to get there.

Re the second question I already tried to explain in #56 this is the same problem as the first. Not just the same method of solution, the very same solution, same situation.

I have to go out and will have to defer my promised rewriting again.

 

[Moose100]

OK I am posting the pics of my notes. This text isn't helping me get this across this is for the first problem 16.32

At "part 3"i did use a polynomial calculator. Where one of the roots is x=-0.032. I wanted to make sure that you saw I was heeding your advice.

Also the values in the mass action ratio reflect me multiplying the equilibrium values by 2 to reflect a doubling in pressure that comes from moving in the the piston halfway.

 

[epenguin]

Seen that - as far as I can see it is what you already said. The authorities frown on use of I-pad photos on the homework forum, but if they must be used presentability can be cleaned up with apps like DocScan HD. Nowever whatever the presentation, realize that someone else's arithmetic is about as inviting as someone else's unmade bed. That's why a minimum of comment in your strings of arithmetic would help readers (examiners) to follow what you're doing. Then if you make a mistake and get wrong answer they can still follow your thought and give you some credit, otherwise they may not try. Also without it you may not understand your calculations yourself when you review them in revision. What I called explicitness. (This also helps you see what relevant information you have, and may suggest solution steps.)

I have set out the calculation in this spirit below. Using Tex has slowed me down so I have only done the first part (comments outside what you'd need to say in italics). For the second you are still invited to get the algebraic formula for p(O₂ for totally solving the seconda part in general; if you really can't do it I will when I complete.

1. Homework Statement
SO3 decomposes on heating:
2SO2(g)

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?

Homework Equations

p(SO_2) = 0.152p(SO_3) \quad...(1) \quad given
K_{eq} = \frac{[SO_2]^2[O_2]}{[SO_3]^2} \quad...(2) \quad equilibrium \quad law

Pressure ∝ Molarity. ...(3)

We could convert if we knew the temperature, but it was not given.
Hence
K = \frac{p(SO_2)^2p(O_2)}{p(SO_3)^2} \quad...(4)

As already discussed this is the same K really as before, but numerically different because expressed in different units.

p(SO_3) + p(SO_2) + p(O_2) = p_{tot} = 2.75 \quad atm \quad...(5)
Dalton law pp.

Calculation

p(SO_2) = 2p(O_2) \quad...(6)
because we start from pure SO₃ and SO₂ and O₂ molecules are formed from it in 2:1 ratio. If instead we started from a general situation, e.g. from mixing arbitrary amounts of oxygen with sulfur oxides we would need to know initial amounts of each and get more complicated conservation equations in two or three 'independent variables'.

Using this (5) becomes
p(SO_3) + \frac{3}{2}p(SO_2) = 2.75 \quad...(7)
Eliminate [SO₃] between this and (1) and get
(\frac{3}{2} + \frac{1}{0.152})p(SO_2) = 2.75 \quad...(8)
Ending with p(SO_2) = 0.34 and so p(O_2) = 0.17 \quad.(9)
pO₂ is all we need, as we already have the ratio (1). (However we can use the p(SO₂ we now have now and the p(SO₃) we can obtain from the additive Dalton law as a check.)
K = 0.152^2×0.17 = 0.00393 \quad atm.

 

[Moose100]

Thank you.

I will see about that equation.

 

[Moose100]

Just posted sometihng but it as wrong! GRRR! This problem is killing me!

Outside of what I have shown (pO2=0.34-2x) I can't see the other way to solve.

 

[epenguin]

Thank you.

I will see about that equation.

Just posted sometihng but it as wrong! GRRR! This problem is killing me!

Outside of what I have shown (pO2=0.34-2x) I can't see the other way to solve.

Schaum books have answers. I asked before - did you get the exact same answer as in the book, or just nearly?

 

[Moose100]

Same thing for both problems

 

[Moose100]

Look at the snapshots. I'm not sure what you're asking about that equation but I've tried can u just show it to me now ? I solved the problem days ago asked you about another and you're still asking me to do this.
Stop telling me how many times you repeated yourself I've tried everything you asked me for almost two weeks. What your doing is egoistic overkill st this pount.

 

[Drakkith]

Yet again, thread locked for moderation.

 

[jedishrfu]

I think the thread has run its course now and so we're closing it with a big thanks to all of you who contributed.

 

[epenguin]

Concluding the arguments of #72 we develop first an equation for the more usual type of problem, where an initial concentration c of SO₃ is given and the reaction is in fixed volume. As reaction proceeds by conservation of sulphur atoms

[SO_3] + [SO_2] = c …(10)

a constant. Then putting together eqns. (2), (10) and (6) we have

\frac{4[O_2]^3}{(c – 3[O_2])^2} …(11)

We will call O₂ x, and this can be written

K(c – 3x)^2 – 4x^3=0 \quad …(12)

In the above case we started from a given total concentration (which would not change as the reaction proceeds at constant volume). In the given problem we are given something slightly different- the total pressure (which does change as reaction proceeds) at final equilibrium. So

K = \frac{4p(O_2)^3}{(p(SO_2))^2}

= \frac{4p(O_2)^3}{ (p(SO_2))^2}

= \frac{4p(O_2)^3}{ (p(SO_2))^2}

= \frac{4p(O_2)^3}{ (p(SO_2) + p(SO_3) + p(O_2) – (p(SO_3) + p(O_2))) ^2}

On the one hand this is

K=\frac{4(p(O_2))^3}{(p_{tot}–3p(O_2))^2} \quad …(13)

But it is better to write this in terms of the invariant

(p(SO_2) + p(SO_3)) = C = p_{tot} – p(O_2) = 2.561

,in this case, as

\frac{4p(O_2)^3}{(C – 2p(O_2))^2}

and representing p(O₂) as x gives the cubic

K(C – 2x)^2 – 4x^3=0 \quad …(14)Then when the system is compressed to half volume it is the C (and nothing else) that is doubled. There is nothing to do but solve the equation with C = 2 × 2.561 by numerically by computer. I get x = 0.273. [O₂] has not doubled but it is not far off. This small deviation from linearity will cause only less than a fifth of such deviation in the total pressure, so that may account for why the OP finds everything doubled. It suggests there is a practical sufficient approximate treatment, but I will leave that to anyone who wants. Maybe later I will post a graph.This is explicitness, for the avoidance of doubts, perhaps excessive for an exam answer, but it is just my opinion that the OP will help himself if he addresses whatever seems to stop him writing out chemical-algebraic equations like #11, 12, etc. and that his right answers would get less credit and goodwill from examiners who are looking for method than another student who got the wrong answer but who just wrote the equations out – apart from the help from equations for actually doing the calculations and knowing and showing what you are doing.