Answer: The concentration of [tex]NO_{2}[/tex] in the given equilibrium mixture is 0.0545 M.
Explanation:
The ratio of concentration of products and reactants raised to the power of their coefficients is called equilibrium constant. The symbol used to denote equilibrium constant is [tex]K_{eq}[/tex].
As the given reaction equation is as follows.
[tex]2NO_{2}(g) \rightleftharpoons N_{2}O_{4}(g)[/tex]
The expression for equilibrium constant of this reaction is as follows.
[tex]K_{eq} = \frac{[N_{2}O_{4}]}{[NO_{2}]^{2}}[/tex]
Now, substitute the given values into above formula as follows.
[tex]K_{eq} = \frac{[N_{2}O_{4}]}{[NO_{2}]^{2}}\\67.3 = \frac{(0.2)}{[NO_{2}]^{2}}\\So, [NO] = \sqrt{\frac{0.2}{67.3}}\\= 0.0545 M[/tex]
Thus, we can conclude that the concentration of [tex]NO_{2}[/tex] in the given equilibrium mixture is 0.0545 M.
