At equilibrium, the concentrations in this system were found to be [n2]=[o2]=0.200 m and [no]=0.400 m.
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?

The concentration in equilibrium of NO is 0,550M.

Explanation:

For the reaction:

N₂(g) + O₂(g) ⇄ 2NO

The equilibrium constant is defined as:

k = [NO]² / [N₂][O₂] (1)

Replacing for the concentrations in equilibrium:

k = (0,400M)² / (0,200M)(0,200M)

k = 4,000

If you add more NO until 0,700M, the equilibrium concentrations will be:

[NO] = 0,700M-2x

[N₂] = 0,200M+x

[O₂] = 0,200M+x

Replacing in (1)

4,000 =  (0,700M-2x)² / (0,200M+x)²

4,000 =  4x²- 2,8x + 0,49 / x² + 0,4x + 0,04

4x² + 1,6x + 0,16 = 4x²- 2,8x + 0,49

4,4x = 0,33

x = 0,075M

That means that concentration in equilibrium of NO is:

[NO] = 0,700M - 2×0,075M = 0,550M

I hope it helps!

atomic mass = 39

number of neutrons = 20

number of protons= 19

electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

the number of electron in first shell = 2

number of electrons in n=3 shell = 8