Chemical Forums
Chemistry Forums for Students => Inorganic Chemistry Forum => Topic started by: chehemi123 on November 30, 2012, 01:58:47 PM
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hey guys,
I have got the following problem:
I want to calculate the partial pressure of (SO3), (SO2) and (O2).
The reaction: SO2 + O2 --> SO3
The pressure p is 1,02bar. The temperature is 873K.
Kp is 65,1.
I came to this equations:
Kp= p2(SO3)/p2(SO2) · p(O2)
and
p=p(SO3) + p(SO2) + p(O2)
How can I calculate the partial pressure of SO2, SO3 and O2 ?
I think there must be another equation. Because there are 3 variables.
Maybe another condition by the temperature?
Can you help me?
Thanks in advance !
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Do you know the starting point of the system? All reactants or all products, perhaps?
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Do you know the starting point of the system? All reactants or all products, perhaps?
And further if all reactants than you need the ratio of SO2 : O2 that you start with.
As stated the problem is under-specified.
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yes:
it is said: there is more or less no SO3 and the part by volume of SO2 is 10%.
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Ah - then you know the starting partial pressures of SO2 and O2, and that will let you write an equation relating the three concentrations at any point during the reaction.
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how can i calculate the starting partial pressures? ???
do i need the starting (total) pressure?
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I was assuming that the pressure and temperature you gave were constants during the reaction
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I am not quite sure....
this is the task:
SO2 is produced by sulphur and O2. This gas mixture is more or less without SO3. The part by volume of SO2 is about 10%. This mixture is conducted through a kiln. There the equilibrium is maintained by the current pressure of 1,02 bar and 873K.
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i hope you can understand me. I am from switzerland actually and studying in America!
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is the ratio of SO2 : O2 the same throughout the whole reaction? or is it changing? because 1mol O2 and 2mol SO2 react...so the ratio actually should change?!
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Assume 1 gmol feed basis
Let x be moles of O2 reacted
2 SO2 + O2 --> 2 SO3
Before: 0.1 0.9 0
IN: 0.1-2x 0.9-x 2x
Total moles 1-x
Mole frac.
[tex]
x_{SO_2}=\frac{0.1-2x}{1-x} \\
x_{O_2}=\frac{0.9-x}{1-x} \\
x_{SO_3}=\frac{2x}{1-x} \\
[/tex]
[tex]
P_{SO_2}=\frac{(0.1-2x)}{(1-x)}P \\
P_{O_2}=\frac{(0.9-x)}{(1-x)}P \\
P_{SO_3}=\frac{2x}{(1-x)}P \\
[/tex]
[tex]
K_p= \frac{P^2_{SO_3}}{ P^2_{SO_2} · P_{O_2} } \\
K_p=\frac{ 4x^2 . (1-x) .P }{ (0.1-2x)^2 . (0.9-x) } \\
[/tex]
P = 1.02bar. Kp = 65.1
Now you have only x as a variable. Solve.
[tex]
\frac{1.02}{65.1} = \frac{ 4x^2 . (1-x) }{ (0.1-2x)^2 . (0.9-x) }
[/tex]
That's a cubic; you ought to get three solutions.
Use x = 0.005305; find a good reason to throw the rest away..[x seems a bit too small; Only 5% of the SO2 reacting away; makes me a bit suspicious I've some bug in my solution. ]
Leave it to you to calculate the partial pressures.
Now, you can hope that I've not messed up my arithmetic.....But I'm pretty sure I have (especially since I already found and corrected 3 mistakes; there's probably more). So redo it yourself! ;D
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i hope you can understand me. I am from switzerland actually and studying in America!
The pressure p is 1,02bar. Kp is 65,1.
Now that you've Americanized yourself, might as well get rid of the habit of using swapped punctuation. ;D
OK, I'm KIDDING. But it might confuse a few people on this side of the bathtub. ;)
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you're a legend ! :) Thanks a lot! I am going to redo it and will tell you if there´s a mistake ;)
Now that you've Americanized yourself, might as well get rid of the habit of using swapped punctuation. ;D
OK, I'm KIDDING. But it might confuse a few people on this side of the bathtub. ;)
I will do my best and try to get rid of it ;D
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I was assuming that the pressure and temperature you gave were constants during the reaction
It would be nice if they were indeed constant. But I don't think that assumption is needed.
In fact, I'm still confused why the T was provided at all. Unless I am making a mistake in my solution?
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maybe the temperature is only there for confusing? :P
but i am not quite sure really...
And there is no pressure and temperature ( at the start ) provided...
I was just redoing the calculation:
is it:
[tex]
K_p= \frac{P^2_{SO_3}}{ P^2_{SO_2} · P_{O_2} } \\
K_p=\frac{ 4x^2 . (1-x) .P }{ (0.1-2x)^2 . (0.9-x) } \\
[/tex]
my solution was:
K_p=\frac{ 4x^2 . (1-x) }{ (0.1-2x)^2 . (0.9-x) .P } \\
[/tex]
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ah sorry...I meant that P is in the denominator ?
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ah sorry...I meant that P is in the denominator ?
You are right.
[tex]
1.02*65.1 = \frac{ 4x^2 . (1-x) }{ (0.1-2x)^2 . (0.9-x) }
[/tex]
What does that give; x=0.044?
It'd mean ~88% conversion on SO2. That sounds like a way more reasonable answer.
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I am not that good in solving these cubics...
I might find a program which can calculate x :P
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I am not that good in solving these cubics...
I might find a program which can calculate x :P
Use Brute Force and Bisection. 0<x<0.05 is not a big range.
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I got
x1= 0,0442599
x2= 0,0574578
x3= 0,898282
x2 and x3 are thrown away ( cause X(So2) would be negative...) ;)
wooop :D
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oh, by the way there is another little problem.
If I have the parts by substance amount of SO2, O2 and SO3 at the equilibrium at 1000K and 1,013*10^5 Pa.
Can I calculate Kp?
Kx= x2(SO3)/x2(SO2)*x(O2)
I found:
Kx= Kp · (p/pt)2
with p/pt as a pressure factor?
is this way right?
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What's p and pt
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I don´t really know ;D
I read it somewhere in the internet...
but shouldn´t be right..
Do you know any relation between Kp and Kx ? o0
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found it:
https://docs.google.com/viewer?a=v&q=cache:7g96HWjj1RIJ:www.psci305.utep.edu/ch1305/Chapter12/gc12.pdf+relation+Kx+Kp&hl=de&gl=de&pid=bl&srcid=ADGEESjpvGVc4ZUnGO_SQ99s13qurl5o3svtmwFTLjMNgfyHgJaNvGUbHi67awMB7G28-nKq_HwB6sDnWGY8PYeqYEAN0IsBE5bBjYeHJUvEq0dxIPfqK09jjNaxdApHGUF05iIhwnKG&sig=AHIEtbS8pBDI7-qGLKYOUUg06vUSHZXSNQ
(page 7)
should be:
Kp=Kx · ptotΔn
hope so :P