Respuesta :
Answer:- Equilibrium partial pressure of NO is 0.50 atm and equilibrium partial pressure of [tex]NO_2[/tex] = 0.02 atm.
Solution:- Let's say the initial pressure of [tex]NO_2[/tex] is p. We would make the ICE table for the given equation and consider the change in pressure as X.
[tex]2NO_2 \leftrightarrow 2NO + O_2[/tex]
I P 0 0
C -2X +2X +X
E (P-2X) 2X X
From given information, equilibrium partial pressure of oxygen is 0.25 atm and from ICE table it is X.
So, X = 0.25
equilibrium partial pressure of NO = 2(0.25 atm) = 0.50 atm
and equilibrium partial pressure of nitrogen dioxide = P-0.50
For the equation we have, the Kc expression would be written as:
[tex]Kp=(\frac{[NO]^2[O_2]}{[NO_2]^2})[/tex]
Let's plug in the values in this expression:
[tex]158=\frac{(0.50)^2(0.25)}{(P-0.50)^2}[/tex]
[tex]158=\frac{0.0625}{(P-0.50)^2}[/tex]
This could also be written as:
[tex](P-0.50)^2=\frac{0.0625}{158}[/tex]
On taking square root to both sides:
(P-0.50) = 0.0199
add 0.50 to both sides:
P = 0.0199 + 0.50
P = 0.5199 and it's rounded off to 0.52
Now, we could calculate the equilibrium partial pressure of nitrogen dioxide as:
equilibrium partial pressure of nitrogen dioxide = P-0.50
equilibrium partial pressure of nitrogen dioxide = 0.52-0.50 = 0.02 atm
So, equilibrium partial pressure of [tex]NO_2[/tex] = 0.02 atm
equilibrium partial pressure of NO = 0.50 atm and
