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chemical

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KP calculation
« on: May 07, 2005, 10:33:07 AM »
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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Re:KP calculation
« Reply #1 on: May 07, 2005, 05:30:08 PM »
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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chemical

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Re:KP calculation
« Reply #2 on: May 07, 2005, 06:10:08 PM »
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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Offline Borek

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Re:KP calculation
« Reply #3 on: May 07, 2005, 08:35:02 PM »
Quote from: chemical on May 07, 2005, 06:10:08 PM
and at equilbrium they wer 0.01  0.4975   0.4975

Not 0.01 0.495 0.495?
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chemical

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Re:KP calculation
« Reply #4 on: May 08, 2005, 07:09:33 AM »
Quote
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)
-------------------------
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
« Last Edit: May 08, 2005, 07:10:33 AM by chemical »
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Re:KP calculation
« Reply #5 on: May 08, 2005, 07:38:14 AM »
Quote from: chemical on May 08, 2005, 07:09:33 AM
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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chemical

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Re:KP calculation
« Reply #6 on: May 08, 2005, 07:57:47 AM »
Quote from: Borek on May 08, 2005, 07:38:14 AM
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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Offline Borek

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Re:KP calculation
« Reply #7 on: May 08, 2005, 09:11:02 AM »
Quote from: chemical on May 08, 2005, 07:57:47 AM
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 ;)

Quote
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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chemical

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Re:KP calculation
« Reply #8 on: May 08, 2005, 10:52:22 AM »
Quote from: Borek on May 08, 2005, 09:11:02 AM
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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chemical

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Re:KP calculation
« Reply #9 on: May 09, 2005, 06:40:16 AM »
my answer is correct?
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Offline AWK

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Re:KP calculation
« Reply #10 on: May 09, 2005, 09:56:48 AM »
No, it should be (0.495)^2/(0.01)^2
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AWK

huh?

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Re:KP calculation
« Reply #11 on: May 20, 2005, 01:19:38 PM »
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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Re:KP calculation
« Reply #12 on: May 20, 2005, 04:47:58 PM »
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
« Last Edit: May 23, 2005, 08:08:05 PM by geodome »
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Offline Borek

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Re:KP calculation
« Reply #13 on: May 22, 2005, 02:56:59 PM »
Quote from: geodome on May 20, 2005, 04:47:58 PM
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?
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