Chemical Forums
Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: chemical on May 07, 2005, 10:33:07 AM
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hi, i,m a newbie to the forum as you can see have been searching for a chemistry forum for a while now and this seems pretty active and good compared to others. :)
ok lets see how smart this place is i have worked out a KP calcuation for the following reaction but i am not too sure if its correct, so could some1 try and answer the following question
at 1600 oC and 1.5 atm pressure NO is 99% dissociated at equilbrium. calculate the value of Kp under these condition.
2NO <------> N2 + O2 all in gase phase
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this is a relatively easy question, but we still adhere to the forum policy that you must show that you have the tried the problem
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hi, ok this is what i did to get my answer:
i got 66.7 (no units).
the starting moles i did 2moles of NO, and zero for both N2 and O2
and at equilbrium they wer 0.01 0.4975 0.4975
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and at equilbrium they wer 0.01 0.4975 0.4975
Not 0.01 0.495 0.495?
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Not 0.01 0.495 0.495?
NO scrap what i said before i did a mistake i should have assumed there was 1 mole rather then write 2 moles.
at equilbrium there wil be NO2 = 0.01 and N2=0.2475 O2= 0.2475
total moles= 0.505
mole fraction: NO2= 0.0198 N2= 0.4901 O2= 0.4901
Pa at 1.5 atm= 0.0132 0.3267 0.3267
p(0.3267)*p(0.3267)
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p(0.0132)2 =612.6 (no units)
pleaaaaaaase could you check if this answer is correct.
May i get a scooby snack if i am correct ;D
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NO scrap what i said before i did a mistake i should have assumed there was 1 mole rather then write 2 moles.
In this case everything cancels out and equlibrium constant can be calculated using either partial pressures or numbers of moles as they are directly proportional.
Start with 1 mole as you did at first. You will have 0.01 mole NO left at equilibrium. Now you have to carefully calculate amount of N2 and O2 at equilibrium.
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In this case everything cancels out and equlibrium constant can be calculated using either partial pressures or numbers of moles as they are directly proportional.
Start with 1 mole as you did at first. You will have 0.01 mole NO left at equilibrium. Now you have to carefully calculate amount of N2 and O2 at equilibrium.
hmmm should i get a new calculator i keep getting wrong calculation answers i had this 1 for over 7 years lol.
well this is what i did 0.01/2= 0.005/2= 2.5 X 10-3
therefore the answer would be 0.063 (no units)
correct??
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hmmm should i get a new calculator i keep getting wrong calculation answers i had this 1 for over 7 years lol.
For such simple questions I am using calculator that is over forty now ;)
well this is what i did 0.01/2= 0.005/2= 2.5 X 10-3
No. If you have started with 1 mole NO and there is 0.01 mole NO left, how much moles of N2 and O2 have beed produced?
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For such simple questions I am using calculator that is over forty now ;)No. If you have started with 1 mole NO and there is 0.01 mole NO left, how much moles of N2 and O2 have beed produced?
0.99/2 = 0.495
but i though when it said 0.99 disociated it means that this amount didnt react?
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my answer is correct?
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No, it should be (0.495)^2/(0.01)^2
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hi,
i dont understand how you came to that answer, this is what i did.
2NO<--> N2 + 02 TOTAL
intial 1.00 0 0
equilbrium
(99%dissociated) 0.01 4.67X10^-3 5.34 X10^-3 0.02
mole fraction 0.01/0.02=0.5 0.2335 0.267
partial pressure at
1.5 atm 1.5x0.5= 0.75 0.35025 0.4005
KP= 0.140275125/ 1.5
kp= 0.09 (NO UNITS)
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let N be initial number of moles of NO
given we have 99% dissociation,
mole of NO = 0.01N
mole of N2 = 0.99N/2 = 0.495N
mole of O2 = 0.99N/2 = 0.495N
total number of moles = 1.00N
using Dalton's Law of Partial Pressure,
partial pressure of N2 = 0.495N/(1.00N) * 1.5atm = 0.7425atm
partial pressure of O2 = 0.495N/(1.00N) * 1.5atm = 0.74125atm
partial pressure of NO = 0.01N/1.00N * 1.5atm = 0.015atm
Kp = (0.7425atm)2/(0.015atm)2 = 2450
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let N be initial number of moles of NO
given we have 99% dissociation,
we have 0.01N of NO, 0.99N of N2 and 0.99N of O2
Have you taken into account the fact that O2 and N2 are diatomic particles?