Respuesta :
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.
