Answer : The correct option is, (E) 7.8 atm
Explanation :
The partial pressure of
= 8.00 atm
The partial pressure of
= 5.00 atm
= 0.0025
The balanced equilibrium reaction is,
![N_2(g)+O_2(g)\rightleftharpoons 2NO(g)](https://tex.z-dn.net/?f=N_2%28g%29%2BO_2%28g%29%5Crightleftharpoons%202NO%28g%29)
Initial pressure 8.00 5.00 0
At eqm. (8.00-x) (5.00-x) 2x
The expression of equilibrium constant
for the reaction will be:
![K_p=\frac{(p_{NO})^2}{(p_{N_2})(p_{O_2})}](https://tex.z-dn.net/?f=K_p%3D%5Cfrac%7B%28p_%7BNO%7D%29%5E2%7D%7B%28p_%7BN_2%7D%29%28p_%7BO_2%7D%29%7D)
Now put all the values in this expression, we get :
![0.0025=\frac{(2x)^2}{(8.00-x)\times (5.00-x)}](https://tex.z-dn.net/?f=0.0025%3D%5Cfrac%7B%282x%29%5E2%7D%7B%288.00-x%29%5Ctimes%20%285.00-x%29%7D)
By solving the terms, we get:
![x=0.15atm](https://tex.z-dn.net/?f=x%3D0.15atm)
The equilibrium partial pressure of
= (8.00 - x) = (8.00 - 0.15) = 7.8 atm
Therefore, the equilibrium partial pressure of
is 7.8 atm.