[chehemi123]

ah sorry...I meant that P is in the denominator ?

[curiouscat]

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.

[chehemi123]

I am not that good in solving these cubics...

I might find a program which can calculate x 

[curiouscat]

I am not that good in solving these cubics...

I might find a program which can calculate x 

Use Brute Force and Bisection. 0<x<0.05 is not a big range.

[chehemi123]

I got

x1= 0,0442599
x2= 0,0574578
x3= 0,898282

x2 and x3 are thrown away ( cause X(So2) would be negative...)

wooop 

[chehemi123]

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?

K_x= x²(SO₃)/x²(SO₂)*x(O₂)

I found:
K_x= K_p · (p/p_t)²

with p/p_t as a pressure factor?

is this way right?

[curiouscat]

What's p and p_t

[chehemi123]

I don´t really know 
I read it somewhere in the internet...
but shouldn´t be right..

Do you know any relation between Kp and Kx ? o0

[chehemi123]

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 · p_tot^Δn

hope so