To determine the equilibrium constant \( K_{\text{eq}} \) for the reaction
[tex]\[ 2 HF (g) \longleftrightarrow H_2(g) + F_2(g) \][/tex]
at 600 K, given the equilibrium concentrations of the reactants and products, we can use the expression for the equilibrium constant for a gaseous reaction:
[tex]\[ K_{\text{eq}} = \frac{[H_2][F_2]}{[HF]^2} \][/tex]
Where:
- \([HF] = 5.82 \times 10^{-2} \text{ M}\)
- \([H_2] = 8.4 \times 10^{-3} \text{ M}\)
- \([F_2] = 8.4 \times 10^{-3} \text{ M}\)
Now, let's plug these concentrations into the equilibrium constant expression:
[tex]\[ K_{\text{eq}} = \frac{(8.4 \times 10^{-3})(8.4 \times 10^{-3})}{(5.82 \times 10^{-2})^2} \][/tex]
Calculating the numerator:
[tex]\[ (8.4 \times 10^{-3}) \times (8.4 \times 10^{-3}) = 8.4^2 \times 10^{-6} = 70.56 \times 10^{-6} = 7.056 \times 10^{-5} \][/tex]
Calculating the denominator:
[tex]\[ (5.82 \times 10^{-2})^2 = 5.82^2 \times 10^{-4} = 33.8724 \times 10^{-4} = 3.38724 \times 10^{-3} \][/tex]
Now, dividing the numerator by the denominator:
[tex]\[ K_{\text{eq}} = \frac{7.056 \times 10^{-5}}{3.38724 \times 10^{-3}} \][/tex]
[tex]\[ K_{\text{eq}} = \frac{7.056}{3.38724} \times 10^{-5+3} \][/tex]
[tex]\[ K_{\text{eq}} \approx 2.0831 \times 10^{-2} \][/tex]
So, the calculated value of \( K_{\text{eq}} \) is approximately \( 2.0831 \times 10^{-2} \).
Among the given options, the value that matches closely is:
[tex]\[ 2.1 \times 10^{-2} \][/tex]
Therefore, the value of \( K_{\text{eq}} \) for the reaction at 600 K is:
[tex]\[ 2.1 \times 10^{-2} \][/tex]
