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Offline chehemi123

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calculating partial pressures
« on: November 30, 2012, 01:58:47 PM »
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 !

Offline fledarmus

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Re: calculating partial pressures
« Reply #1 on: November 30, 2012, 03:08:17 PM »
Do you know the starting point of the system? All reactants or all products, perhaps?

Offline curiouscat

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Re: calculating partial pressures
« Reply #2 on: November 30, 2012, 03:11:03 PM »
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.

Offline chehemi123

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Re: calculating partial pressures
« Reply #3 on: November 30, 2012, 04:08:56 PM »
yes:

it is said: there is more or less no SO3 and the part by volume of SO2 is 10%.

Offline fledarmus

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Re: calculating partial pressures
« Reply #4 on: November 30, 2012, 04:12:19 PM »
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.

Offline chehemi123

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Re: calculating partial pressures
« Reply #5 on: November 30, 2012, 04:23:43 PM »
how can i calculate the starting partial pressures?  ???
do i need the starting (total) pressure?

Offline fledarmus

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Re: calculating partial pressures
« Reply #6 on: November 30, 2012, 04:31:11 PM »
I was assuming that the pressure and temperature you gave were constants during the reaction

Offline chehemi123

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Re: calculating partial pressures
« Reply #7 on: November 30, 2012, 04:54:13 PM »
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.


Offline chehemi123

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Re: calculating partial pressures
« Reply #8 on: December 01, 2012, 05:46:26 AM »
i hope you can understand me. I am from switzerland actually and studying in America!

Offline chehemi123

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Re: calculating partial pressures
« Reply #9 on: December 01, 2012, 06:17:33 AM »
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?!

Offline curiouscat

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Re: calculating partial pressures
« Reply #10 on: December 01, 2012, 07:26:02 AM »
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
« Last Edit: December 01, 2012, 08:20:55 AM by curiouscat »

Offline curiouscat

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Re: calculating partial pressures
« Reply #11 on: December 01, 2012, 07:58:20 AM »
i hope you can understand me. I am from switzerland actually and studying in America!

Quote
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.  ;)

Offline chehemi123

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Re: calculating partial pressures
« Reply #12 on: December 01, 2012, 11:40:21 AM »
you're a legend !  :)  Thanks a lot! I am going to redo it and will tell you if there´s a mistake  ;)

Quote

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

Offline curiouscat

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Re: calculating partial pressures
« Reply #13 on: December 01, 2012, 12:04:45 PM »
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?

Offline chehemi123

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Re: calculating partial pressures
« Reply #14 on: December 01, 2012, 12:25:18 PM »
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:

Quote
[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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