To determine the equilibrium constant [tex]\( K_{eq} \)[/tex] for the reaction:
[tex]\[ 2 H_2(g) + S_2(g) \rightleftharpoons 2 H_2S(g) \][/tex]
we follow these steps:
1. Calculate the concentrations of each species in the 1.00-liter vessel.
- The concentration of [tex]\( H_2 \)[/tex]:
[tex]\[
[H_2] = \frac{\text{moles of } H_2}{\text{volume of vessel}} = \frac{0.50 \, \text{moles}}{1.00 \, \text{liter}} = 0.50 \, \text{M}
\][/tex]
- The concentration of [tex]\( S_2 \)[/tex]:
[tex]\[
[S_2] = \frac{\text{moles of } S_2}{\text{volume of vessel}} = \frac{0.020 \, \text{moles}}{1.00 \, \text{liter}} = 0.020 \, \text{M}
\][/tex]
- The concentration of [tex]\( H_2S \)[/tex]:
[tex]\[
[H_2S] = \frac{\text{moles of } H_2S}{\text{volume of vessel}} = \frac{68.5 \, \text{moles}}{1.00 \, \text{liter}} = 68.5 \, \text{M}
\][/tex]
2. Write the equilibrium expression for the reaction:
[tex]\[
K_{eq} = \frac{[H_2S]^2}{[H_2]^2 \cdot [S_2]}
\][/tex]
3. Substitute the equilibrium concentrations into the equilibrium expression:
- [tex]\([H_2] = 0.50 \, \text{M}\)[/tex]
- [tex]\([S_2] = 0.020 \, \text{M}\)[/tex]
- [tex]\([H_2S] = 68.5 \, \text{M}\)[/tex]
Therefore, the equilibrium constant is:
[tex]\[
K_{eq} = \frac{(68.5)^2}{(0.50)^2 \cdot 0.020}
\][/tex]
4. Calculate the numerical value:
[tex]\[
K_{eq} = \frac{(68.5 \times 68.5)}{(0.50 \times 0.50) \times 0.020}
\][/tex]
[tex]\[
K_{eq} = \frac{4692.25}{0.25 \times 0.020}
\][/tex]
[tex]\[
K_{eq} = \frac{4692.25}{0.005}
\][/tex]
[tex]\[
K_{eq} = 938450.0
\][/tex]
So, the equilibrium constant [tex]\( K_{eq} \)[/tex] for this reaction system is [tex]\( 9.4 \times 10^5 \)[/tex].
Thus, the correct option is:
[tex]\[ K = 9.4 \times 10^5 \][/tex]
