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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: o1ocups on February 16, 2009, 02:32:39 AM

Title: Another problem about predicting the pressure-am I on the right track?
Post by: o1ocups on February 16, 2009, 02:32:39 AM

Question

At a given temperature a system containing  and some oxides of nitrogen can be described by the following reactions:
2NO(g) + O₂(g) ::equil:: 2No₂(g) K_p=10⁴
2NO₂(g) ::equil:: N2O4(g) K_p=0.10
A pressure of 1 atm of N₂O₄(g) is placed in a container at this temperature.

Predict which, if any component (other than ) will be present at a pressure greater than 0.2 atm at equilibrium.
A) The pressures of NO and O₂ will be negligible.
B) NO will be present.
C) NO and O₂ will be present.
D) O₂ will be present.

What I attempted

2NO(g) + O₂(g) ::equil:: 2NO₂(g) Kp=10⁴
2NO₂(g) ::equil:: N2O4(g) K_p=0.10

2NO(g) + O₂(g) ::equil:: N2O4(g)
10₄*0.10=10000=K_p

K_p=P_N2O4/P_O2P_NO

P_O2P_NO=K_p/P_N2O4

P_O2P_NO=1/10000=negligible??

I am not sure what I am doing. Is this right?  ???

Title: Re: Another problem about predicting the pressure-am I on the right track?
Post by: Astrokel on February 16, 2009, 10:57:38 AM

Quote

2NO(g) + O2(g)  N2O4(g)
104*0.10=10000=Kp

ICE table. What do you see after putting it into your K_p expression?

Title: Re: Another problem about predicting the pressure-am I on the right track?
Post by: o1ocups on February 16, 2009, 01:58:33 PM

	NO	O2	N2O4
I	?	?	1 atm
C
E	0.2??	0.2??

I don't understand what you mean  :-[
Sorry, I am really slow.

Title: Re: Another problem about predicting the pressure-am I on the right track?
Post by: Astrokel on February 16, 2009, 05:17:21 PM

It's ok, take it slowly as long as you give it a try. Always follow the question and it says you only have 1 atm of N₂O₄, so assumption is NO and O₂ is present at 0 initially. Ok so start working with dissociation next.

    2NO  O2  ::equil:: N2O4
I    0    0          1 atm
C  +2x  +x        -x
E

i guess you know what to do next and just observe what you get.

Title: Re: Another problem about predicting the pressure-am I on the right track?
Post by: o1ocups on February 16, 2009, 06:46:14 PM

Thank you SO MUCH for your time and energy!! I got the right answer!  :D
I just find the information about the initial conditions really ambiguous. I never know what to assume  :-\
But thanks again!!

Title: Re: Another problem about predicting the pressure-am I on the right track?
Post by: pandabear on February 16, 2009, 07:49:42 PM

i still don't understand what the answer is...sorry, i'm an idiot.

Title: Re: Another problem about predicting the pressure-am I on the right track?
Post by: o1ocups on February 16, 2009, 08:31:48 PM

Quote from: pandabear on February 16, 2009, 07:49:42 PM

i still don't understand what the answer is...sorry, i'm an idiot.

I'll give you a hint: how do you calculate Kp?