# Equilibrium Question

Discussion in 'MCAT Study Question Q&A' started by HopefulOncoDoc, Jan 10, 2010.

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1. ### HopefulOncoDocNew Member

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Could someone please explain why choice C is the correct answer? Thanks!

If a flask were filled with pure CO2 (g) to a total pressure of 1.00 atm., then once equilibrium is reached, the total pressure of the system is which of the following?

2CO (g) + O2 (g) <--> 2CO2 (g)

A. Less than 1.00 atm.
B. Exactly 1.00 atm.
C. Between 1.00 atm. and 1.50 atm.
D. Exactly 1.50 atm.
Last edited: Jan 11, 2010
2. ### boazshanah alef

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Is there something missing from the question? E.g. temp change, water etc.
3. ### HopefulOncoDocNew Member

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Yes sorry about that I totally forgot. I put the rxn up above. Here is the table from the passage..

Temperature (K).....Kp (atm^-1)
375 .....................9
425 .....................20.8
500 .....................32.3
600 .....................60.5

This is from TBR Gchem Section 3 Equilibrium Passage I Question 5. I just couldn't understand their explaination.
4. ### loveoforganic-Account Deactivated-

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So you have 100% product, at 1.0 atm of product. If that were to all convert to reactant, you'd have 1.5 atm of reactant. It's an equlibrium though, so you shift to some mix of reactant and product, which must put you between 1.0 and 1.5 atm.
5. ### HopefulOncoDocNew Member

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Thanks for the response. Could you explain how you would get 1.5 atm. if that were to all convert to reactant? I understand why the pressure would increase since there's more reactants on the side..
6. ### Econ2MDNew Member

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Pure CO2 will be converted to CO and O2 via the reverse reaction. Since all are in the gaseous state, pure CO2 is 2 moles of molecules, CO and O2 is three moles of molecules. The sum of their partial pressures at equilibrium exceeds that of CO2 alone, so the pressure at equilibrium will exceed the pressure of CO2 alone but be less than the pressure of CO and O2 alone (3 moles). PV=nRT where P is the sum of the partial pressures. The nature of this relationship is independent of temperature.
7. ### Econ2MDNew Member

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2 moles of molecules corresponds to 1 atmosphere of pressure, so 3 moles of molecules corresponds to 1.5 atm pressure. PV=nRT or P=nRT/V. Increase n by 1.5x, increases P by 1.5x also.
8. ### HopefulOncoDocNew Member

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Makes sense. Thanks a lot!
9. ### ilovemediSDN Two Year Member

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oh god. i've been stuck at this problem FOREVER. can ANYONE provide ANY insight? . how did they get total pressure?
10. ### milski1K member

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#4 in this thread has the answer. You start with the equilibrium fully to the right. If you push it all the way left, you'll get 3/2 more moles of gas and 3/2 more pressure. Since the equilibrium is somewhere between these two extremes, it has to be between 1 and 1.5 atm.
11. ### ThirdEyeNew Member

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You have 1atm, of pure product. For every 2atm of prod you lose, you gain 3atm of reactants. If you lost all product (1atm), you would gain all reactant (1.5 atm). You will NEVER lose all prod because it's an eq rxn, so you can't quite get up to 1.5atm.

This problem is way simpler than it appears
12. ### ilovemediSDN Two Year Member

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So basically, 2/1 = 3/x ; and x = 1.5 but since it doesn't go fully to the right (you have some products), it's less than 1.5. Makes sense.

But then for passage 3 - SIMILAR concept:
PCl3 + Cl2 ---> PCl5
If a flask were filled with PCl3 and Cl2 to a total pressure of 1.5atm such that the mole fraction of PCl3 is twice that of Cl2, then wha is the total pressure of the system at equilibrium?

A) Less than 1.00atm
B)between 1 and 1.25 atm
c) between 1.25 and 1.5 atm
d) greater than 1.5 atm.

Ok so after some calculations, we know that PCl3= 1 atm; Cl2 = .5 atm initially. I did the same thing - 2/1.5=1/x and got .75 as my "max" product it an form, but that's obviously not a choice. I also tried the ICE box thing that was in the answer.. Isn't this suppose to be like the question above ^^?

How do you approach this problem?
Last edited: Jul 2, 2012
13. ### californiamedNew Member

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Also do not get... Anyone know how to do passage 3 in last post?
14. ### ThirdEyeNew Member

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The most that can react is .5 atm of each reactant because they are 1:1 and you only have .5 atm of cl2

.5 and .5 (1 atm total) react yielding .5 atm of product with .5 atm of PCl3 remaining.

.5 product + .5 left over pcl3 = 1
15. ### milski1K member

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That's what I did as well, but why is then the maximum 1.25 and not 1.5? What if the equilibrium is just a tiny bit away from the initial condition?
16. ### ThirdEyeNew Member

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Maximum is 1, not 1.25.

You start with 1.5 and gain .5 for every 1 you lose. You can only lose a max of 1 total (.5 each) because of the limiting reagent.

1.5 - 1 + .5 = 1
17. ### ilovemediSDN Two Year Member

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This may seem dumb, but - I know you start w/ 1.5 total pressure, but how'd you know you "gain .5 for every 1 you lose". Gain product?

Also you say ".5 and .5 (1 atm total) react" - but isnt' there 1 of PCl3 and .5 of Cl2? So 1 and .5 react..?
18. ### ThirdEyeNew Member

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Don't worry about sounding dumb, this stuff used to confuse the crap out of me so I can relate.

You have 2 moles of reactants(1 mole of each) that yield 1 mole of product. This is the same as saying 1 mole of reactants will yield .5 moles of prod. So 1 atm reactants becomes .5 atm prod. Again, these are totals.

The reactants react together in a 1 to 1 ratio. Since you only have .5 atm of cl2 (because one is 2x the other), only .5 atm of pcl3 can react because ALL of the cl2 gets used up and pcl3 now has nothing to react with
19. ### ilovemediSDN Two Year Member

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Thanks, makes sense. So this problem is different from the last one I was talking about because the CO2 started out with PRODUCT first and no reactants were stated (thus, we couldn't use the 'limit reagent' trick')?
20. ### milski1K member

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Ok, still not making any sense. I understand how we get 1 (which is the minimum, not the maximum?) but the right answer is that B, which says that the equilibrium partial pressure is between 1 and 1.25. Where does 1.25 come from?
21. ### ThirdEyeNew Member

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1 is actually the min and the max. It's not a double headed arrow so it's assumed to go all the way. If it were a double headed arrow, we'd need the Keq to figure it out but it would still most likely fall between 1 and 1.25 because in this case 1 would still be the min but not the max.

The answer choice says it falls between 1 and 1.25. 1 DOES fall between 1 and 1.25. Watch out for tricky wording
22. ### ThirdEyeNew Member

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Sort of. If you had more than one product, they would need to come back together to form the reactant(s) and in this case one would be limiting if there were less molar equivalents of it.
23. ### milski1K member

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Got it, I missed the arrow and was trying to figure out the whole range from all the way left to all the way right. Thanks!
24. ### ilovemediSDN Two Year Member

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Thanks third eye and milkski for your help! Still a weird question for me, and I pray I won't receive something like this. Until then I'll do a lot of equilibrium passages..