[chehemi123]

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 !

[fledarmus]

Do you know the starting point of the system? All reactants or all products, perhaps?

[curiouscat]

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.

[chehemi123]

yes:

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

[fledarmus]

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.

[chehemi123]

how can i calculate the starting partial pressures? 
do i need the starting (total) pressure?

[fledarmus]

I was assuming that the pressure and temperature you gave were constants during the reaction

[chehemi123]

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.

[chehemi123]

i hope you can understand me. I am from switzerland actually and studying in America!

[chehemi123]

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?!

[curiouscat]

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! 

[curiouscat]

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. 

OK, I'm KIDDING. But it might confuse a few people on this side of the bathtub. 

[chehemi123]

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. 

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 

[curiouscat]

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?

[chehemi123]

maybe the temperature is only there for confusing?
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]