To determine the equilibrium constant [tex]\( K_{\text{eq}} \)[/tex] for the reaction [tex]\( N_2 + 3H_2 \rightleftharpoons 2NH_3 \)[/tex], we use the expression for the equilibrium constant in terms of the concentrations of the reactants and products at equilibrium.
The equilibrium constant expression for the given reaction is:
[tex]\[ K_{\text{eq}} = \frac{[NH_3]^2}{[N_2] \cdot [H_2]^3} \][/tex]
Given the equilibrium concentrations:
- [tex]\([NH_3] = 3 \, M\)[/tex]
- [tex]\([N_2] = 2 \, M\)[/tex]
- [tex]\([H_2] = 1 \, M\)[/tex]
We can plug these values into the equilibrium constant expression:
[tex]\[ K_{\text{eq}} = \frac{(3 \, M)^2}{(2 \, M) \cdot (1 \, M)^3} \][/tex]
Calculate the numerator and the denominator separately:
- Numerator: [tex]\((3 \, M)^2 = 9 \, M^2\)[/tex]
- Denominator: [tex]\( (2 \, M) \cdot (1 \, M)^3 = 2 \, M \cdot 1 \, M^3 = 2 \, M^4 \)[/tex]
Now, divide the numerator by the denominator to get [tex]\( K_{\text{eq}} \)[/tex]:
[tex]\[ K_{\text{eq}} = \frac{9 \, M^2}{2 \, M^4} = \frac{9}{2} = 4.5 \][/tex]
Thus, the equilibrium constant [tex]\( K_{\text{eq}} \)[/tex] for the reaction is:
[tex]\[ \boxed{4.5} \][/tex]
So the correct answer is:
D. [tex]\( K_{\text{eq}} = 4.5 \)[/tex]
