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Topic: Equilibrium Partial Pressure  (Read 8737 times)

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Offline A5HLEY

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Equilibrium Partial Pressure
« on: February 28, 2008, 10:37:18 AM »
Hi all. I'm having some trouble with this problem. Any suggestions would be greatly appreciated!

At 900 K the following reaction has Kp=0.345:
2SO2(g)+O2(g) <==> 2SO3(g)

In an equilibrium mixture the partial pressures of SO2 and O2 are .165 atm and .755 atm, respectively. What is the equilibrium partial pressure of SO3 in the mixture?

Ok, so I'm pretty sure that Kp=P[SO3]^2/P[SO2]^2*P[O2]

So do I just sub in those values and solve for P of SO3, or am I missing the big picture?

Thanks for your *delete me*

Online Borek

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Re: Equilibrium Partial Pressure
« Reply #1 on: February 28, 2008, 10:59:35 AM »
So do I just sub in those values and solve for P of SO3

Looks like :)
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Offline AWK

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Re: Equilibrium Partial Pressure
« Reply #2 on: February 28, 2008, 11:53:39 AM »
Quote
Kp=P[SO3]^2/P[SO2]^2*P[O2]

This is denominator - you shoud use brackets
(P[SO2]^2*P[O2])
AWK

Offline A5HLEY

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Re: Equilibrium Partial Pressure
« Reply #3 on: February 29, 2008, 07:37:50 PM »
Ok, so will someone tell me if this is correct?

I derived the formula Kp=[SO3]^2/([SO2]2[O2]), and I got [SO3]^2=Kp*[SO2]^2*[O2]

So I plugged in the numbers and got [SO3]^2=(.345)(.755)(.165^2), and got that [SO3]^2=.0070914319, took the square root and got .0842106399.

Is that right?

Offline A5HLEY

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Re: Equilibrium Partial Pressure
« Reply #4 on: March 01, 2008, 12:30:58 PM »
Anyone? :)

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