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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: mandy9008 on February 20, 2009, 06:41:42 PM

Title: Partial Pressures
Post by: mandy9008 on February 20, 2009, 06:41:42 PM

2NO(g) + Br₂(g)  ::equil:: 2NOBr(g)

K_P=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?

Title: Re: Partial Pressures
Post by: Astrokel on February 20, 2009, 11:34:41 PM

Hmm am i missing something. What is your K_p unit?

Title: Re: Partial Pressures
Post by: azure 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.

Title: Re: Partial Pressures
Post by: rabolisk on February 05, 2011, 05:39:50 PM

Are you sure what you were given was K_p?

Title: Re: Partial Pressures
Post by: Jorriss 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.

Title: Re: Partial Pressures
Post by: DevaDevil on February 08, 2011, 10:36:25 AM

Quote from: Jorriss on February 08, 2011, 01:51:57 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.

Quote from: azure on February 05, 2011, 05:23:10 PM

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

yes it does

Quote from: azure on February 05, 2011, 05:23:10 PM

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 K_p.
There is also no temperature given, which would make this unsolvable.


K_p = p_NOBr² / (p_NO² * p_Br2 )

the problem with this is indeed that they did not give any unit with K_p, 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?

Title: Re: Partial Pressures
Post by: ok123 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.)

Title: Re: Partial Pressures
Post by: mjc123 on October 01, 2019, 12:41:08 PM

K_p has no units if each pressure is expressed as a multiple of the standard pressure, i.e.

K_p = (P_NOBr/P°)²/{(P_NO/P°)²*P_Br2/P°}

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