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Topic: calculating partial pressures  (Read 9757 times)

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Offline chehemi123

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Re: calculating partial pressures
« Reply #15 on: December 01, 2012, 12:27:14 PM »
ah sorry...I meant that P is in the denominator ?

Offline curiouscat

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Re: calculating partial pressures
« Reply #16 on: December 01, 2012, 12:42:20 PM »
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.

Offline chehemi123

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Re: calculating partial pressures
« Reply #17 on: December 01, 2012, 01:18:20 PM »
I am not that good in solving these cubics...

I might find a program which can calculate x  :P

Offline curiouscat

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Re: calculating partial pressures
« Reply #18 on: December 01, 2012, 01:51:59 PM »
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.

Offline chehemi123

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Re: calculating partial pressures
« Reply #19 on: December 01, 2012, 01:55:45 PM »
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

Offline chehemi123

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Re: calculating partial pressures
« Reply #20 on: December 01, 2012, 02:08:57 PM »
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?

Offline curiouscat

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Re: calculating partial pressures
« Reply #21 on: December 01, 2012, 02:19:55 PM »
What's p and pt

Offline chehemi123

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Re: calculating partial pressures
« Reply #22 on: December 01, 2012, 02:25:46 PM »
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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