If the concentration equilibrium constant for this reaction is Kc = 3.36 at reaction conditions, what will the equilibrium concentration of NO(g) be if a sealed reaction vessel is filled with gases at initial concentrations of 0.0 M of NO3(g), 1.8 M of NO(g), and 2.8 M of NO2(g)?

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sebassandin sebassandin · Answer 1 · 2019-11-04T19:53:17+01:00

Answer:

[tex][NO]_{eq}=0.329M[/tex]

Explanation:

Hello,

The balanced chemical reaction must be stated as follows:

[tex]NO(g)+NO_2(g)-->N_2O_3(g)[/tex]

Thus, the equilibrium constant, thanks to the law of mass action turns out being:

[tex]Kc=\frac{[N_2O_3]_{eq}}{[NO]_{eq}[NO_2]_{eq}}[/tex]

Based on the equilibrium condition, one leave the law of mass action in terms of the change [tex]x[/tex] due to the chemical reaction:

[tex]Kc=\frac{x}{(1.8M-x)(2.8M-x)}\\(5.04-4.6x+x^2)*3.36=x\\16.9344-15.456x-x+3.36x^2=0\\3.36x^2-16.456x+16.9344=0\\x_1=1.471M\\x_2=3.427M[/tex]

The adequate result is 1.471M since the other one turns out into a negative molarity for NO.

Finally, one conclude that the equilibrium concentration of NO is:

[tex][NO]_{eq}=1.8M-1.471M=0.329M[/tex]

Best regards.