To find the equilibrium constant ([tex]\( K_c \)[/tex]) for the reaction [tex]\( N_2(g) + 3 H_2(g) \rightleftharpoons 2 NH_3(g) \)[/tex], we can use the following expression for the equilibrium constant:
[tex]\[
K_c = \frac{[NH_3]^2}{[N_2] \cdot [H_2]^3}
\][/tex]
Given the equilibrium concentrations:
[tex]\[
[NH_3] = 0.105 \, M
\][/tex]
[tex]\[
[N_2] = 1.1 \, M
\][/tex]
[tex]\[
[H_2] = 1.50 \, M
\][/tex]
Now, plug these values into the equilibrium constant expression:
[tex]\[
K_c = \frac{(0.105)^2}{(1.1) \cdot (1.50)^3}
\][/tex]
This evaluates to:
[tex]\[
\frac{0.011025}{1.1 \cdot 3.375}
\][/tex]
First, calculate the denominator:
[tex]\[
1.1 \cdot 3.375 = 3.7125
\][/tex]
Now, complete the division:
[tex]\[
K_c = \frac{0.011025}{3.7125} \approx 0.00297
\][/tex]
Therefore, the equilibrium constant [tex]\( K_c \)[/tex] for this reaction at the given temperature is approximately 0.00297. Among the given choices, the closest value is:
[tex]\[
\boxed{0.0030}
\][/tex]
