To find the equilibrium constant \( K_c \) for the given reaction at equilibrium, we can use the equilibrium concentrations given and the formula for the equilibrium constant for the reaction:
[tex]\[ \text{N}_2(g) + 3 \text{H}_2(g) \longleftrightarrow 2 \text{NH}_3(g) \][/tex]
The equilibrium constant expression for this reaction is:
[tex]\[ K_c = \frac{{[\text{NH}_3]^2}}{{[\text{N}_2] \cdot [\text{H}_2]^3}} \][/tex]
Given:
[tex]\[ [\text{NH}_3] = 0.105 \, M \][/tex]
[tex]\[ [\text{N}_2] = 1.1 \, M \][/tex]
[tex]\[ [\text{H}_2] = 1.50 \, M \][/tex]
We substitute these values into the equilibrium constant expression:
[tex]\[ K_c = \frac{{(0.105)^2}}{{(1.1) \cdot (1.50)^3}} \][/tex]
Evaluating the numerator and the denominator:
[tex]\[ \text{Numerator} = (0.105)^2 = 0.011025 \][/tex]
[tex]\[ \text{Denominator} = 1.1 \cdot (1.50)^3 = 1.1 \cdot 3.375 = 3.7125 \][/tex]
Now, divide the numerator by the denominator:
[tex]\[ K_c = \frac{0.011025}{3.7125} \approx 0.0029697 \][/tex]
Rounding this to three significant figures, we get:
[tex]\[ K_c \approx 0.0030 \][/tex]
Therefore, the equilibrium constant [tex]\( K_c \)[/tex] for the reaction at this temperature is 0.0030.
