Answer:
[tex][N_2]_{eq}=0.094M[/tex]
[tex][O_2]_{eq}=0.094M[/tex]
[tex][NO]_{eq}=0.002M[/tex]
Explanation:
Hello,
In this case, for the given chemical reaction, we write the following law of mass action:
[tex]Keq=\frac{[N_2][O_2]}{[NO]^2}[/tex]
That in terms of the change [tex]x[/tex] due to the reaction extent (ICE procedure) we can write:
[tex]Keq=\frac{x*x}{([NO]_0-2x)^2}[/tex]
Thus, the initial concentration of nitrogen monoxide is:
[tex][NO]_0=\frac{0.570mol}{3.0L} =0.19M[/tex]
Thereby, we write:
[tex]2400=\frac{x*x}{(0.19-2x)^2}[/tex]
That we can solve by suing the quadratic equation formula or solver to obtain two roots:
[tex]x_1=0.094M\\x_2=0.096M[/tex]
Nevertheless, the correct answer is 0.094 M since the other root will produce a negative concentration of nitrogen monoxide at equilibrium, therefore, the equilibrium concentrations turn out:
[tex][N_2]_{eq}=x=0.094M[/tex]
[tex][O_2]_{eq}=x=0.094M[/tex]
[tex][NO]_{eq}=0.19M-2x=0.19M-2(0.094M)=0.002M[/tex]
Best regards.
