Question

At equilibrium, the concentrations in this system were found to be [N2]=[O2]=0.300M and
[NO]=0.500M.

N2(g)+O2(g) −⇀ 2NO(g)

If more NO is added, bringing its concentration to 0.800M, what will the final concentration of NO be after equilibrium is re‑established?

Answer 1

Equilibrium constant K in reaction aA + bB = cC + dD is K = $\tfrac{С^c*D^d}{A^a*B^b}$

in this reaction

K = $\tfrac{[NO]^2}{[N_2]*[O_2]}$ =$\tfrac{[0.5]^2}{[0.3]*[0.3]}$ =$\tfrac{25}{9}$ =2.778

after reestablishing the equilibrium concentrations of O₂ and N₂ will be 0.3+x each and concentration of NO will be 0.8-2x. K doesn't change. Then

K = 2.778= $\tfrac{25}{9}$ = $\tfrac{(0.8-2x)^2}{(0.3+x)*(0.3+x)}$ =$\tfrac{(0.8-2x)^2}{(0.3+x)^2}$

after extracting the square root from both parts

$\tfrac{5}{3}$ = $\tfrac{0.8-2x}{0.3+x}$

1.5+5x = 2.4-6x

0.9=11x

x = $\tfrac{0.9}{11}$ =0.082

[NO]=0.8-2x=0.8-0.082*2=0.636 M