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 01-10-2010, 11:23 PM #1 Member     Status: Pre-Medical Join Date: Mar 2009 Location: CA Posts: 84 Equilibrium Question SDN Members don't see this ad. (About Ads) 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 by HopefulOncoDoc; 01-11-2010 at 04:07 AM.
01-11-2010, 04:02 AM   #2
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Quote:
 Originally Posted by HopefulOncoDoc 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? 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.
Is there something missing from the question? E.g. temp change, water etc.
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01-11-2010, 04:11 AM   #3
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Quote:
 Originally Posted by ishchayill Is there something missing from the question? E.g. temp change, water etc.
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.

 01-11-2010, 06:13 AM #4 -Account Deactivated-   Join Date: Jan 2009 Posts: 4,247 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. __________________ -Account Deactivated-
01-11-2010, 12:45 PM   #5
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Quote:
 Originally Posted by loveoforganic 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.
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..

 01-11-2010, 12:53 PM #6 Senior Member   Status: Medical Student Join Date: Oct 2009 Posts: 142 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.
01-11-2010, 12:55 PM   #7
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Quote:
 Originally Posted by HopefulOncoDoc 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..
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.

01-11-2010, 01:33 PM   #8
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 Originally Posted by Econ2MD 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.
Makes sense. Thanks a lot!

 07-02-2012, 09:11 PM #9 Senior Member   Join Date: Nov 2011 Posts: 206 oh god. i've been stuck at this problem FOREVER. can ANYONE provide ANY insight? . how did they get total pressure?
 07-02-2012, 09:29 PM #10 1K Member     Status: Pre-Medical Join Date: Dec 2009 Location: Where the rain grows Posts: 1,864 #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.
07-02-2012, 09:29 PM   #11
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 Originally Posted by ilovemedi oh god. i've been stuck at this problem FOREVER. can ANYONE provide ANY insight? . how did they get total pressure?
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

 07-02-2012, 09:39 PM #12 Senior Member   Join Date: Nov 2011 Posts: 206 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. Answer (highlight)- B 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 by ilovemedi; 07-02-2012 at 10:45 PM.
 07-02-2012, 10:46 PM #13 Senior Member   Join Date: Jul 2008 Posts: 103 Also do not get... Anyone know how to do passage 3 in last post?
07-02-2012, 11:10 PM   #14
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Quote:
 Originally Posted by ilovemedi 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. Answer (highlight)- B 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?
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

07-02-2012, 11:15 PM   #15
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 Originally Posted by ThirdEye 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
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?

07-02-2012, 11:37 PM   #16
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 Originally Posted by milski 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?
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

07-02-2012, 11:41 PM   #17
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Quote:
 Originally Posted by ThirdEye 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
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..?

07-02-2012, 11:57 PM   #18
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Quote:
 Originally Posted by ilovemedi 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..?
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

 07-03-2012, 12:12 AM #19 Senior Member   Join Date: Nov 2011 Posts: 206 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')?
07-03-2012, 07:05 AM   #20
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 Originally Posted by ThirdEye 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
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?

07-03-2012, 11:55 AM   #21
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Quote:
 Originally Posted by milski 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?
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

07-03-2012, 11:59 AM   #22
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 Originally Posted by ilovemedi 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')?
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.

07-03-2012, 12:04 PM   #23
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 Originally Posted by ThirdEye 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
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!

 07-04-2012, 12:47 AM #24 Senior Member   Join Date: Nov 2011 Posts: 206 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..

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