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### Topic: equilibrium  (Read 2718 times)

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#### orgo814

• Full Member
•    • Posts: 412
• Mole Snacks: +11/-6 ##### equilibrium
« on: April 07, 2012, 08:57:29 PM »
Hey Guys,

These are equilibrium based problems. Any help would be appreciated.

1) For an ICE problem, i got down to the expression Kp = (2x)^2/(0.500 - 3x)^3 (0.900 - 2x) = 4.3 x 10^-4.
Since the K value is small, can I assume the (-3x) and (-2x) is negligible and just cube 0.500 and multiply that by 0.900 to get my denominator value?

2) Here's the problem:
The equilibrium constant Kp of the reaction: 2SO3(g) = 2SO2(g) + O2(g) is 7.69 at 830 celsius. If a vessel at this temperature initially contains pure SO3 and if the partial pressure of SO3 at equilibrium is 0.100 atm, what is the partial pressure of O2 in the flask at equilbrium.

My Kp expression is (PSO2)^2 (PO2) / (PSO3)^2. I know there has to be an initial PSO3... they just didn't tell us. In a previous problem they used A to represent that initial value. The reaction shifts right so my equilibrium expressions are A - 2x for SO3, 2x for SO2, and X for O2. They're telling me that the equilibrium pressure for SO3 is 0.100. So, I assume I'd have to solve for X but I have no idea how since they didn't give me enough information.... If I can solve for X, that x would be my equilibrium pressure for O2.

#### Borek ##### Re: equilibrium
« Reply #1 on: April 08, 2012, 05:05:21 AM »
For an ICE problem, i got down to the expression Kp = (2x)^2/(0.500 - 3x)^3 (0.900 - 2x) = 4.3 x 10^-4.
Since the K value is small, can I assume the (-3x) and (-2x) is negligible and just cube 0.500 and multiply that by 0.900 to get my denominator value?

You can, just afterwards plug calculated x into the Kp and see how accurate the result is.

Quote
The equilibrium constant Kp of the reaction: 2SO3(g) = 2SO2(g) + O2(g) is 7.69 at 830 celsius. If a vessel at this temperature initially contains pure SO3 and if the partial pressure of SO3 at equilibrium is 0.100 atm, what is the partial pressure of O2 in the flask at equilbrium.

My Kp expression is (PSO2)^2 (PO2) / (PSO3)^2. I know there has to be an initial PSO3... they just didn't tell us.

Initially there was just a pure SO3, doesn't it tell you what was the initial PSO3?
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#### orgo814

• Full Member
•    • Posts: 412
• Mole Snacks: +11/-6 ##### Re: equilibrium
« Reply #2 on: April 08, 2012, 06:28:49 PM »
Thanks for the reply. They did not give us any initial pressure for SO3.

#### Borek ##### Re: equilibrium
« Reply #3 on: April 09, 2012, 04:24:45 AM »