Answer to Question #306638 in General Chemistry for Sofia

Question

At equilibrium, the concentrations in this system were found to be[N

2

]=[O

2

]=0.200 M

and[NO]=0.500 M.

N

2

(g)+O

2

(g)↽−

−⇀

2NO(g)

If more NO

 is added, bringing its concentration to 0.800 M,

 what will the final concentration of NO

 be after equilibrium is re‑established?

Answer 1

K_eq = [NO]²/[N₂]*[O₂] = (0.5)²/(0.2*0.2) = 6.25

Let the concentration of N₂ and O₂ increase by x M after after equilibrium is re‑established. 

N₂(g)  +  O₂(g)  =  2NO(g)

0.2+x + 0.2+x = 0.8 - 2x

K_eq = (0.8-2x)²/(0.2 + x)² = 6.25

After solving this the equation we get x = 0.067 M

So final concentration of NO be after equilibrium is re‑established = 0.800 - 2*0.067 = 0.667 M