At equilibrium, the concentrations in this system were found to be [ N 2 ] = [ O 2 ] = 0.100 M [N2]=[O2]=0.100 M and [ NO ] = 0.600 M . [NO]=0.600 M. N 2 ( g ) + O 2 ( g ) − ⇀ ↽ − 2 NO ( g ) N2(g)+O2(g)↽−−⇀2NO(g) If more NO NO is added, bringing its concentration to 0.900 M, 0.900 M, what will the final concentration of NO NO be after equilibrium is re‑established?

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OlaMacgregor OlaMacgregor · Answer 1 · 2020-04-17T10:58:42+02:00

Answer: The final concentration of NO be after equilibrium is re‑established is 0.825 M.

Explanation:

The given balanced chemical equation is as follows.

[tex]N_{2}(g) + O_{2}(g) \rightleftharpoons 2NO(g)[/tex]

Now, equilibrium constant for this reaction will be as follows.

[tex]K_{c} = \frac{[NO]^{2}}{[N_{2}][O_{2}]}[/tex]

It is given that concentrations at the equilibrium are:

[tex][N_{2}] = [O_{2}][/tex] = 0.1 M and [NO] = 0.6 M

Therefore, the value of [tex]K_{c}[/tex] is as follows.

[tex]K_{c} = \frac{(0.6)^{2}}{(0.1)(0.1)}[/tex]

= 36.0

NO concentration of 0.9 M is added to the system. So,

[tex]N_{2}(g) + O_{2}(g) \rightleftharpoons 2NO(g)[/tex]

Initial: 0.1 0.1 0.9

Change: x x 2x

Equibm:(0.1 + x) (0.1 + x) (0.9 - 2x)

Now, we will find the value of x as follows.

36.0 = [tex]\frac{(0.9 - 2x)^{2}}{(0.1 + x)^{2}}[/tex]

x = 0.0375

Therefore, final concentration of NO after the equilibrium that is re-established is as follows.

0.9 - 2x

= [tex]0.9 - (2 \times 0.0375)[/tex]

= 0.9 - 0.075

= 0.825 M

Therefore, we can conclude that the final concentration of NO be after equilibrium is re‑established is 0.825 M.