Question

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?

Answer 1

Answer:

0.84M

Explanation:

Hello,

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:

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

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

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

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

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

Thus, we arrange the equation as:

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

Finally, the new concentration is:

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

Best regards.