Question

A sample of pure NO2 gas decomposes at 1000 K

2NO2 (g) ↔ 2 NO (g) + O2 (g)
  The constant Kp is 158. An analysis shows that the partial pressure of O2 is 0.25 atmospheres at equilibrium. Determine the pressure of NO and NO2.

Answer 1

Answer:- Equilibrium partial pressure of NO is 0.50 atm and equilibrium partial pressure of
`NO_2` = 0.02 atm.

Solution:- Let's say the initial pressure of
`NO_2` is p. We would make the ICE table for the given equation and consider the change in pressure as X.

`2NO_2 \leftrightarrow 2NO + O_2`

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:

`Kp=(([NO]^2[O_2])/([NO_2]^2))`

Let's plug in the values in this expression:

`158=((0.50)^2(0.25))/((P-0.50)^2)`

`158=(0.0625)/((P-0.50)^2)`

This could also be written as:

`(P-0.50)^2=(0.0625)/(158)`

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
`NO_2` = 0.02 atm

equilibrium partial pressure of NO = 0.50 atm and