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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: A5HLEY on February 28, 2008, 10:37:18 AM

Title: Equilibrium Partial Pressure
Post by: A5HLEY 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*

Title: Re: Equilibrium Partial Pressure
Post by: Borek on February 28, 2008, 10:59:35 AM

Quote from: A5HLEY on February 28, 2008, 10:37:18 AM

So do I just sub in those values and solve for P of SO3

Looks like :)

Title: Re: Equilibrium Partial Pressure
Post by: AWK 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])

Title: Re: Equilibrium Partial Pressure
Post by: A5HLEY 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?

Title: Re: Equilibrium Partial Pressure
Post by: A5HLEY on March 01, 2008, 12:30:58 PM

Anyone? :)