Answer :
To determine the equilibrium constant [tex]\( K_{\text{eq}} \)[/tex] for the reaction:
[tex]\[ 2 HF (g) \longleftrightarrow H_2 (g) + F_2 (g), \][/tex]
we use the given equilibrium concentrations. The formula for the equilibrium constant [tex]\( K_{\text{eq}} \)[/tex] for this reaction can be written as:
[tex]\[ K_{\text{eq}} = \frac{[H_2][F_2]}{[HF]^2}. \][/tex]
Given the equilibrium concentrations:
[tex]\[ [HF] = 5.82 \times 10^{-2} \text{ M}, \][/tex]
[tex]\[ [H_2] = 8.4 \times 10^{-3} \text{ M}, \][/tex]
[tex]\[ [F_2] = 8.4 \times 10^{-3} \text{ M}, \][/tex]
we substitute these values into the formula for [tex]\( K_{\text{eq}} \)[/tex]:
[tex]\[ K_{\text{eq}} = \frac{(8.4 \times 10^{-3})(8.4 \times 10^{-3})}{(5.82 \times 10^{-2})^2}. \][/tex]
Now, let's calculate it step-by-step:
1. Calculate the numerator:
[tex]\[ (8.4 \times 10^{-3}) \times (8.4 \times 10^{-3}) = 70.56 \times 10^{-6} = 7.056 \times 10^{-5}. \][/tex]
2. Calculate the denominator:
[tex]\[ (5.82 \times 10^{-2})^2 = 33.8724 \times 10^{-4} = 3.38724 \times 10^{-3}. \][/tex]
3. Divide the numerator by the denominator:
[tex]\[ K_{\text{eq}} = \frac{7.056 \times 10^{-5}}{3.38724 \times 10^{-3}} \approx 2.083 \times 10^{-2}. \][/tex]
Upon rounding to two significant figures, we get:
[tex]\[ K_{\text{eq}} \approx 2.1 \times 10^{-2}. \][/tex]
Thus, the value of [tex]\( K_{\text{eq}} \)[/tex] for the reaction is [tex]\( 2.1 \times 10^{-2} \)[/tex]. Therefore, the correct answer is:
[tex]\[ \boxed{2.1 \times 10^{-2}} \][/tex]
[tex]\[ 2 HF (g) \longleftrightarrow H_2 (g) + F_2 (g), \][/tex]
we use the given equilibrium concentrations. The formula for the equilibrium constant [tex]\( K_{\text{eq}} \)[/tex] for this reaction can be written as:
[tex]\[ K_{\text{eq}} = \frac{[H_2][F_2]}{[HF]^2}. \][/tex]
Given the equilibrium concentrations:
[tex]\[ [HF] = 5.82 \times 10^{-2} \text{ M}, \][/tex]
[tex]\[ [H_2] = 8.4 \times 10^{-3} \text{ M}, \][/tex]
[tex]\[ [F_2] = 8.4 \times 10^{-3} \text{ M}, \][/tex]
we substitute these values into the formula for [tex]\( K_{\text{eq}} \)[/tex]:
[tex]\[ K_{\text{eq}} = \frac{(8.4 \times 10^{-3})(8.4 \times 10^{-3})}{(5.82 \times 10^{-2})^2}. \][/tex]
Now, let's calculate it step-by-step:
1. Calculate the numerator:
[tex]\[ (8.4 \times 10^{-3}) \times (8.4 \times 10^{-3}) = 70.56 \times 10^{-6} = 7.056 \times 10^{-5}. \][/tex]
2. Calculate the denominator:
[tex]\[ (5.82 \times 10^{-2})^2 = 33.8724 \times 10^{-4} = 3.38724 \times 10^{-3}. \][/tex]
3. Divide the numerator by the denominator:
[tex]\[ K_{\text{eq}} = \frac{7.056 \times 10^{-5}}{3.38724 \times 10^{-3}} \approx 2.083 \times 10^{-2}. \][/tex]
Upon rounding to two significant figures, we get:
[tex]\[ K_{\text{eq}} \approx 2.1 \times 10^{-2}. \][/tex]
Thus, the value of [tex]\( K_{\text{eq}} \)[/tex] for the reaction is [tex]\( 2.1 \times 10^{-2} \)[/tex]. Therefore, the correct answer is:
[tex]\[ \boxed{2.1 \times 10^{-2}} \][/tex]