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Topic: Partial Pressures  (Read 16781 times)

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Offline mandy9008

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Partial Pressures
« on: February 20, 2009, 06:41:42 PM »
2NO(g) + Br2(g)  ::equil:: 2NOBr(g)

KP=28.4

in a reaction mixture at equilibrium, the partial pressure of NO is 111 torr and that of Br2 is 183 torr.

What is the partial pressure of NOBr?

what i did:
k=[NOBr]2/[NO]2[Br2]

28.4=x2/(1112)(183)
x=8002.168531torr

the answer is 290 torr.
How do i arrive here?

Offline Astrokel

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Re: Partial Pressures
« Reply #1 on: February 20, 2009, 11:34:41 PM »
Hmm am i missing something. What is your Kp unit?
No matters what results are waiting for us, it's nothing but the DESTINY!!!!!!!!!!!!

Offline azure

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Re: Partial Pressures
« Reply #2 on: February 05, 2011, 05:23:10 PM »
First off, I must address Astrokel. The equilibrium constant does not have units.
Now for the real question, you are forgetting to factor in your temperature, try using this formula to solve it: 28.4= [(x/(RT))^2]/[(111/RT)^2*(183/RT)]
Just do some algebra to the formula, plug in the correct value (and units) for R, and then plug in your temperature.
Note, this may not necessarily work, because I believe this equation actually factors in Kc as well.
« Last Edit: February 05, 2011, 05:35:32 PM by azure »

Offline rabolisk

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Re: Partial Pressures
« Reply #3 on: February 05, 2011, 05:39:50 PM »
Are you sure what you were given was Kp?

Offline Jorriss

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Re: Partial Pressures
« Reply #4 on: February 08, 2011, 01:51:57 AM »
I'm guessing Br2 is suppose to be liquid.

And K does have units in this case. You get conc^2/conc^3, that has units.

Offline DevaDevil

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Re: Partial Pressures
« Reply #5 on: February 08, 2011, 10:36:25 AM »
I'm guessing Br2 is suppose to be liquid.

In the question: " in a reaction mixture at equilibrium, the partial pressure of NO is 111 torr and that of Br2 is 183 torr."
so no, it is a gas here.

First off, I must address Astrokel. The equilibrium constant does not have units.
yes it does


Now for the real question, you are forgetting to factor in your temperature, try using this formula to solve it: 28.4= [(x/(RT))^2]/[(111/RT)^2*(183/RT)]

not with Kp.
There is also no temperature given, which would make this unsolvable.


Kp = pNOBr2 / (pNO2 * pBr2 )

the problem with this is indeed that they did not give any unit with Kp, so you don't know if it is torr-1, Pa-1, atm-1 etc.

based on your given numbers, I do not get to the supposed answer. Can you check the numbers and the question for us again?

Offline ok123

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Re: Partial Pressures
« Reply #6 on: October 01, 2019, 10:14:18 AM »
You must convert torr to atm for the calculation. Then convert back to torr at the end. (I suppose Kp is based off pressure in atm and that's why.)

Offline mjc123

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Re: Partial Pressures
« Reply #7 on: October 01, 2019, 12:41:08 PM »
Kp has no units if each pressure is expressed as a multiple of the standard pressure, i.e.

Kp = (PNOBr/P°)2/{(PNO/P°)2*PBr2/P°}

If the standard pressure is 1 atm, the pressure values in torr must be divided by 760.


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