Answer :
We start by finding the initial concentration of [tex]$\mathrm{NO_2}$[/tex] in the container. With 0.800 moles of [tex]$\mathrm{NO_2}$[/tex] in a 5.00 L container, the initial concentration is
[tex]$$
[\mathrm{NO_2}]_{\text{initial}} = \frac{0.800 \text{ mol}}{5.00 \text{ L}} = 0.16 \text{ M}.
$$[/tex]
The reaction is
[tex]$$
2\,\mathrm{NO_2} \rightleftharpoons \mathrm{N_2O_4}.
$$[/tex]
We are told that at equilibrium the concentration of [tex]$\mathrm{N_2O_4}$[/tex] is 0.066 M. Let [tex]$x = 0.066$[/tex] M be the amount of [tex]$\mathrm{N_2O_4}$[/tex] formed at equilibrium.
Since 2 moles of [tex]$\mathrm{NO_2}$[/tex] are consumed for every mole of [tex]$\mathrm{N_2O_4}$[/tex] formed, the decrease in [tex]$\mathrm{NO_2}$[/tex] is
[tex]$$
\Delta [\mathrm{NO_2}] = 2x = 2(0.066 \text{ M}) = 0.132 \text{ M}.
$$[/tex]
Thus, the equilibrium concentration of [tex]$\mathrm{NO_2}$[/tex] is
[tex]$$
[\mathrm{NO_2}]_{\text{eq}} = 0.16 \text{ M} - 0.132 \text{ M} = 0.028 \text{ M}.
$$[/tex]
The equilibrium constant expression for the reaction is
[tex]$$
K_c = \frac{[\mathrm{N_2O_4}]}{[\mathrm{NO_2}]^2}.
$$[/tex]
Substitute the equilibrium concentrations into the expression:
[tex]$$
K_c = \frac{0.066 \text{ M}}{(0.028 \text{ M})^2}.
$$[/tex]
Calculating the denominator:
[tex]$$
(0.028 \text{ M})^2 = 0.000784 \text{ M}^2.
$$[/tex]
Now, compute [tex]$K_c$[/tex]:
[tex]$$
K_c = \frac{0.066 \text{ M}}{0.000784 \text{ M}^2} \approx 84.18.
$$[/tex]
Thus, the equilibrium constant for the reaction is
[tex]$$
K_c \approx 84.2.
$$[/tex]
[tex]$$
[\mathrm{NO_2}]_{\text{initial}} = \frac{0.800 \text{ mol}}{5.00 \text{ L}} = 0.16 \text{ M}.
$$[/tex]
The reaction is
[tex]$$
2\,\mathrm{NO_2} \rightleftharpoons \mathrm{N_2O_4}.
$$[/tex]
We are told that at equilibrium the concentration of [tex]$\mathrm{N_2O_4}$[/tex] is 0.066 M. Let [tex]$x = 0.066$[/tex] M be the amount of [tex]$\mathrm{N_2O_4}$[/tex] formed at equilibrium.
Since 2 moles of [tex]$\mathrm{NO_2}$[/tex] are consumed for every mole of [tex]$\mathrm{N_2O_4}$[/tex] formed, the decrease in [tex]$\mathrm{NO_2}$[/tex] is
[tex]$$
\Delta [\mathrm{NO_2}] = 2x = 2(0.066 \text{ M}) = 0.132 \text{ M}.
$$[/tex]
Thus, the equilibrium concentration of [tex]$\mathrm{NO_2}$[/tex] is
[tex]$$
[\mathrm{NO_2}]_{\text{eq}} = 0.16 \text{ M} - 0.132 \text{ M} = 0.028 \text{ M}.
$$[/tex]
The equilibrium constant expression for the reaction is
[tex]$$
K_c = \frac{[\mathrm{N_2O_4}]}{[\mathrm{NO_2}]^2}.
$$[/tex]
Substitute the equilibrium concentrations into the expression:
[tex]$$
K_c = \frac{0.066 \text{ M}}{(0.028 \text{ M})^2}.
$$[/tex]
Calculating the denominator:
[tex]$$
(0.028 \text{ M})^2 = 0.000784 \text{ M}^2.
$$[/tex]
Now, compute [tex]$K_c$[/tex]:
[tex]$$
K_c = \frac{0.066 \text{ M}}{0.000784 \text{ M}^2} \approx 84.18.
$$[/tex]
Thus, the equilibrium constant for the reaction is
[tex]$$
K_c \approx 84.2.
$$[/tex]