Sure, let's solve this step-by-step.
The balanced chemical equation for the reaction is:
[tex]\[ \text{N}_2 + 3\text{H}_2 \leftrightarrow 2\text{NH}_3 \][/tex]
The equilibrium constant expression for this reaction [tex]\( K_{eq} \)[/tex] is defined as:
[tex]\[ K_{eq} = \frac{[\text{NH}_3]^2}{[\text{N}_2] \cdot [\text{H}_2]^3} \][/tex]
We are given the following equilibrium concentrations:
[tex]\[ [\text{NH}_3] = 0.105 \, M \][/tex]
[tex]\[ [\text{N}_2] = 1.1 \, M \][/tex]
[tex]\[ [\text{H}_2] = 1.50 \, M \][/tex]
Now, let's substitute these concentrations into the equilibrium constant expression:
[tex]\[ K_{eq} = \frac{(0.105)^2}{(1.1) \cdot (1.50)^3} \][/tex]
First, calculate the numerator:
[tex]\[ (0.105)^2 = 0.011025 \][/tex]
Next, calculate the denominator:
[tex]\[ (1.50)^3 = 1.50 \times 1.50 \times 1.50 = 3.375 \][/tex]
[tex]\[ 1.1 \times 3.375 = 3.7125 \][/tex]
Now, divide the numerator by the denominator:
[tex]\[ K_{eq} = \frac{0.011025}{3.7125} \approx 0.002969696969696969 \][/tex]
Looking at the provided choices, the closest value to our calculation is [tex]\( 0.0030 \)[/tex].
Therefore, the equilibrium constant [tex]\( K_{eq} \)[/tex] for the reaction at this temperature is:
[tex]\[ \boxed{0.0030} \][/tex]
