Answer :
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]
[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]