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]
