At equilibrium, the concentrations in this system were found to be [N2]=[O2]=0.200 M and [NO]=0.600 M. N2(g)+O2(g)↽−−⇀2NO(g) If more NO is added, bringing its concentration to 0.900 M, what will the final concentration of NO be after equilibrium is re‑established?

At first, the equilibrium constant should be computed because the whole situation is at the same temperature so it is suitable for the new condition, thus:

[tex]K_{eq}=\frac{[NO]^2_{eq}}{[N_2]_{eq}[O_2]_{eq}} \\K_{eq}=\frac{0.6^2}{0.2*0.2}\\ K_{eq}=9[/tex]

Now, the new equilibrium condition, taking into account the change x, becomes:

[tex]9=\frac{[NO]^2_{eq}}{[N_2]_{eq}[O_2]_{eq}}\\9=\frac{[0.9+2x]^2}{[0.2-x][0.2-x]}[/tex]

Nevertheless, since the addition of NO implies that the equilibrium is leftward shifted, we should change the equilibrium constant the other way around:

[tex]\frac{1}{9} =\frac{[N_2]_{eq}[O_2]_{eq}}{[NO]^2_{eq}}\\\frac{1}{9} =\frac{[0.2+x][0.2+x]}{[0.9-2x]^2}[/tex]

Thus, we arrange the equation as:

[tex]\frac{1}{9} (0.9-2x)^2=(0.2+x)^2\\0.09-0.4x+4x^2=0.04+0.4x+x^2\\3x^2-0.8x+0.05=0\\x_1=0.06[/tex]

Finally, the new concentration is:

[tex][NO]_{eq}=0.9-0.06=0.84M[/tex]

Best regards.