Consider the following reaction involving the production of dinitrogen tetroxide:

\[ 2 \text{NO}_{2(g)} \rightleftharpoons \text{N}_2\text{O}_{4(g)} \quad \Delta H = 98.2 \, \text{kJ} \]

0.800 mol of $\text{NO}_2$ is injected into a highly pressured sealed 5.00 L container and allowed to reach equilibrium. At equilibrium, 0.066 M of $\text{N}_2\text{O}_4$ is present. What is the $K_c$ value?

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]