Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started 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) + O_{2}(g) ::equil:: 2No_{2}(g) K_{p}=10^{4}
2NO_{2}(g) ::equil:: N2O4(g) K_{p}=0.10
A pressure of 1 atm of N_{2}O_{4}(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 O_{2} will be negligible.
B) NO will be present.
C) NO and O_{2} will be present.
D) O_{2} will be present.
What I attempted
2NO(g) + O_{2}(g) ::equil:: 2NO_{2}(g) Kp=10^{4}
2NO_{2}(g) ::equil:: N2O4(g) K_{p}=0.10
2NO(g) + O_{2}(g) ::equil:: N2O4(g)
10_{4}*0.10=10000=K_{p}
K_{p}=P_{N2O4}/P_{O2}P_{NO}
P_{O2}P_{NO}=K_{p}/P_{N2O4}
P_{O2}P_{NO}=1/10000=negligible??
I am not sure what I am doing. Is this right? ???

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?

 NO  O2  N2O4 
I  ?  ?  1 atm 
C    
E  0.2??  0.2??  
I don't understand what you mean :[
Sorry, I am really slow.

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 N_{2}O_{4}, so assumption is NO and O_{2} 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.

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!!

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

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?