# At equilibrium, the concentrations in this system were found to be [N2]=[O2]=0.200 M[N2]=[O2]=0.200 M and [NO]=0.500 M.[NO]=0.500 M.N2(g)+O2(g)−⇀↽−2NO(g)N2(g)+O2(g)↽−−⇀2NO(g)If more NONO is added, bringing its concentration to 0.800 M,0.800 M, what will the final concentration of NONO be after equilibrium is re‑established?

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At equilibrium, the concentrations in this system were found to be [N2]=[O2]=0.200 M[N2]=[O2]=0.200 M and [NO]=0.500 M.[NO]=0.500 M.

N2(g)+O2(g)−⇀↽−2NO(g)N2(g)+O2(g)↽−−⇀2NO(g)

If more NONO is added, bringing its concentration to 0.800 M,0.800 M, what will the final concentration of NONO be after equilibrium is re‑established?

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Step 1

The equilibrium constant, Kc for a reaction is given as : Step 2

Given,

Equilibrium concentration of N2 = 0.2 M

Equilibrium concentration of O2 = 0.2 M

Equilibrium concentration of NO = 0.5 M

The balanced reaction is given as :

N2 (g) + O2 (g) ↔ 2 NO (g)    K = ?

The equilibrium constant, K can be calculated as : Step 3

More NO is added, bringing its concentration to 0.800 M.

Now, since NO is added to the equilibrium reaction, to re-establish the equilibrium, the reaction will now move in backward direction according to Le-Chatelier’s principle, so given reaction becomes :

2 NO (g) ↔ N2 (g) + O2 (g)     K’ = 1/K = 1/6.2...

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