Given that the initial concentration of NO is 0.175 M and using the previous analysis, we can determine the equilibrium concentration of $\text{N}_2$ based on the reaction:

$2 \text{NO}(g) \rightleftharpoons \text{N}_2(g) + \text{O}_2(g)$

From the ICE table set up earlier:

Species	Initial (M)	Change (M)	Equilibrium (M)
$\text{NO}$	0.175	$-2x$	$0.175 - 2x$
$\text{N}_2$	0	$+x$	$x$
$\text{O}_2$	0	$+x$	$x$

We calculated earlier that $x$, the change in concentration, is approximately $0.0866 \, M$.

Equilibrium concentration of $\text{N}_2$:

$[\text{N}_2]_{eq} = x = 0.0866 \, M$

So, the equilibrium concentration of $\text{N}_2$ is approximately $0.0866 \, M$.

Do you want further explanation or have any other questions?

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Here are some related questions to consider:

1. How would the equilibrium concentration of $\text{N}_2$ change if the initial concentration of NO were increased?
2. What is the effect of temperature on the equilibrium concentration of $\text{N}_2$?
3. How can you calculate the partial pressure of $\text{N}_2$ at equilibrium?
4. What is the relationship between the equilibrium constant $K_c$ and the concentrations of reactants and products?
5. How does the stoichiometry of the reaction affect the equilibrium concentrations?
6. What would happen to the equilibrium concentration of $\text{N}_2$ if the volume of the system is decreased?
7. How would the equilibrium position change if a catalyst is added to the reaction?
8. Can the equilibrium concentration of $\text{N}_2$ be directly measured experimentally?

Tip: When dealing with equilibrium problems, carefully track the stoichiometric coefficients since they directly influence how concentration changes relate to each other.