At equilibrium, the concentrations in this system were found to be [N2]=[O2]=0.200 M and [NO]=0.400 M.

Question

N2(g)+O2(g)↽−−⇀2NO(g)

If more NO is added, bringing its concentration to, 0.700 M, what will the final concentration of NO be after equilibrium is re‑established?

Answer 1

Let’s calculate the equilibrium constant:

Keq = [NO]²/[N₂][O₂] = (0.4)² /(0.2)(0.2) = 4

Now, look at what happens when NO (product) is added @equilb.

N₂ + O₂ <===> 2NO

0.2.....0.2............0.7...I

+x.....+x..............0.7-2x...C

0.2+x..0.2+x......0.7-x....E

Keq = 4 = (0.7-x)² / (0.2+x)²

Solve for x and the [NO] after re-establishing equilibrium will be 0.7 - x.