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.
  • #71
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.
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20150920_174543[1].jpg
 
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  • #72
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)
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?

Homework Equations



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

Pressure ∝ Molarity. ...(3)

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

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

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

Calculation

[tex]p(SO_2) = 2p(O_2) \quad...(6)[/tex]
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
[tex]p(SO_3) + \frac{3}{2}p(SO_2) = 2.75 \quad...(7)[/tex]
Eliminate [SO3] between this and (1) and get
[tex](\frac{3}{2} + \frac{1}{0.152})p(SO_2) = 2.75 \quad...(8)[/tex]
Ending with [tex]p(SO_2) = 0.34[/tex] and so [tex]p(O_2) = 0.17 \quad.(9)[/tex]
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.)
[tex]K = 0.152^2×0.17 = 0.00393 \quad atm.[/tex]
 
  • #73
Thank you.

I will see about that equation.
 
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  • #74
Just posted sometihng but it as wrong! GRRR! This problem is killing me!:frown::mad:

Outside of what I have shown (pO2=0.34-2x) I can't see the other way to solve.
 
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  • #75
Moose100 said:
Thank you.

I will see about that equation.

Moose100 said:
Just posted sometihng but it as wrong! GRRR! This problem is killing me!:frown::mad:

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?
 
  • #76
Same thing for both problems
 
  • #77
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.
 
  • #78
Yet again, thread locked for moderation.
 
  • #79
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.
 
  • #80
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

[tex] [SO_3] + [SO_2] = c …(10) [/tex]

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

[tex] \frac{4[O_2]^3}{(c – 3[O_2])^2} …(11)[/tex]

We will call O2 x, and this can be written

[tex] K(c – 3x)^2 – 4x^3=0 \quad …(12) [/tex]

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

[tex] K = \frac{4p(O_2)^3}{(p(SO_2))^2}[/tex]

[tex] = \frac{4p(O_2)^3}{ (p(SO_2))^2}[/tex]

[tex] = \frac{4p(O_2)^3}{ (p(SO_2))^2}[/tex]

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

On the one hand this is

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

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

[tex](p(SO_2) + p(SO_3)) = C = p_{tot} – p(O_2) = 2.561 [/tex]

,in this case, as

[tex]\frac{4p(O_2)^3}{(C – 2p(O_2))^2}[/tex]

and representing p(O2) as x gives the cubic

[tex] K(C – 2x)^2 – 4x^3=0 \quad …(14) [/tex]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.
 
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