toxin
Harmless
Posts: 27
Registered: 12-8-2005
Member Is Offline
Mood: No Mood
|
|
Equilibrium Problem
The equilibrium constant (Kp) for 2NO(g) <---> N2(g) + O2(g) is 2,400 at 200 °C
Suppose a closed vessel is charged with 36.1 atm of NO. At equilibrium, how could I find the partial pressure of O2 ?
|
|
|
The_Davster
A pnictogen
Posts: 2861
Registered: 18-11-2003
Member Is Offline
Mood: .
|
|
first you must use a variable, say 'x' for the changes in concentration as the system is going towards equilibrium. You then make an "ICE" table to
determine the equilibrium concentrations of each in terms of x and 36.1. Then you must setup your Kp equation with the equilibrium concentrations,
then you will get a quatratic equation which must be solved with the quadratic formula. Then you got your partial pressure of O2.
Try working through it first and I'll check your answer...
|
|
|
toxin
Harmless
Posts: 27
Registered: 12-8-2005
Member Is Offline
Mood: No Mood
|
|
Thanks for your responce, do you mean like this ?
_________2NO(g) <---> N2(g) + O2(g)
Initial__P__| 36.1___|___0___|__0___|
Eq____P__| 36.1-x__|___0+x_|__0+x_|
Kp= (PaN2)*(PaO2) / (PaNO)^2 = (x^2) / ((36.1-x)^2) = 2,400
rewriting this as x^2 - 2,400*(36.1 - x)*(36.1 - x) = -2,399*x^2 + 173,280*x -3,127,704 = 0 and then solving for x gives x=35.37 or 36.85 but 35.37 is
the only choice that makes sense, and the closest choice I have to that is 35.7 is that correct ?
[Edited on 13-3-2006 by toxin]
[Edited on 13-3-2006 by toxin]
|
|
|
The_Davster
A pnictogen
Posts: 2861
Registered: 18-11-2003
Member Is Offline
Mood: .
|
|
It should be 36.1-2x, as there are two moles of NO
|
|
|
toxin
Harmless
Posts: 27
Registered: 12-8-2005
Member Is Offline
Mood: No Mood
|
|
Ahh yes I got a much closer answer choice now when I solve for x in 9599*x^2 - 346,560*x + 3,127,704, but now I am faced with a seemingly similar
problem however I am not given the inital partial pressures.
consider the equilibrium NH4HS(s) <---> NH3(g) + H2S(g) in which Kp = 0.07 at 22 °C
A sample of NH4HS is placed in an evacuated container and allowed to come to equilibrium. The partial pressure of NH3 is then increased by the
addition of 0.590 atm of NH3.
How could you find the partial pressure H2S at this new equilibrium considering that according to Le Chatliers principle the system will shift to the
left(in order to achieve a new equilibrium) ?
[Edited on 13-3-2006 by toxin]
[Edited on 13-3-2006 by toxin]
|
|
|
Darkblade48
Hazard to Others
Posts: 411
Registered: 27-3-2005
Location: Canada
Member Is Offline
Mood: No Mood
|
|
I'm actually not sure how to do this problem, it's been too long since I actually had to solve equilbrium problems (i.e. I spent my entire year doing
quantum chemistry and thermodynamics). I feel ashamed 
I took a look at this problem and for some reason, I have two unknowns, but only one equation, and thus I can't solve for a numerical answer for the
new partial pressure of the H2S.
I'm also curious as to the phase of NH4HS, I thought that it should be a gas and not a solid? If it's a solid, it won't contribute to the equilbrium
calculation though...
|
|
|
toxin
Harmless
Posts: 27
Registered: 12-8-2005
Member Is Offline
Mood: No Mood
|
|
do you think that you can equate PaNH3 to PaH2S, does that make any sense ?
Then the problem would be approached by making a table as such, noting that pure solids or liquids are not part of the equilibrium expression
_________NH4HS(s) <---> NH3(g) + H2S(g)
Initial__P__|------------------|_0.59__|__0___|
Eq____P__|-------------------|_0.59-x= PaH2S|
considering Kp= (0.59-x)^2=0.07 you can find x, what do you think ?
[Edited on 13-3-2006 by toxin]
|
|
|
Darkblade48
Hazard to Others
Posts: 411
Registered: 27-3-2005
Location: Canada
Member Is Offline
Mood: No Mood
|
|
The main problem that I see is the fact that the statement "A sample of NH4HS is placed in an evacuated container and allowed to come to equilibrium "
means that there will already be NH3 and H2S in your initial state (i.e. Pa of H2S is not 0 as you have indicated in the initial state).
|
|
|
toxin
Harmless
Posts: 27
Registered: 12-8-2005
Member Is Offline
Mood: No Mood
|
|
I was assuming that it will be equivalent to restate it, as if the system went into equilibrium together with the extra ammonia added instantly.
I finally decided on a set up like this in which x is the unkown ammounts after equilibrium.
p(NH3)*p(H2S) = 0.07 --> p(NH3) = p(H2S) = 0.26
pnew(NH3) = x + 0.590
pnew(NH3) * pnew(H2S) = 0.07
(x+0.590) * x = 0.07=Kp
...and then solve for x but now its too late for me to change my mind anyways
[Edited on 14-3-2006 by toxin]
|
|
|
![[*]](images/xpblue/default_icon.gif){kind=icon}
