Answer : The correct option is, (E) 7.8 atm
Explanation :
The partial pressure of [tex]N_2[/tex] = 8.00 atm
The partial pressure of [tex]O_2[/tex] = 5.00 atm
[tex]K_p[/tex] = 0.0025
The balanced equilibrium reaction is,
[tex]N_2(g)+O_2(g)\rightleftharpoons 2NO(g)[/tex]
Initial pressure 8.00 5.00 0
At eqm. (8.00-x) (5.00-x) 2x
The expression of equilibrium constant [tex]K_p[/tex] for the reaction will be:
[tex]K_p=\frac{(p_{NO})^2}{(p_{N_2})(p_{O_2})}[/tex]
Now put all the values in this expression, we get :
[tex]0.0025=\frac{(2x)^2}{(8.00-x)\times (5.00-x)}[/tex]
By solving the terms, we get:
[tex]x=0.15atm[/tex]
The equilibrium partial pressure of [tex]N_2[/tex] = (8.00 - x) = (8.00 - 0.15) = 7.8 atm
Therefore, the equilibrium partial pressure of [tex]N_2[/tex] is 7.8 atm.