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!

answer: h+ its less than oh

explanation:

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