To determine the equilibrium constant [tex]\( K_c \)[/tex] for the reaction at 600 K, we will use the given equilibrium concentrations. The balanced chemical equation for the reaction is:
[tex]\[
H_2(g) + CO_2(g) \rightarrow H_2O(g) + CO(g)
\][/tex]
The expression for the equilibrium constant [tex]\( K_c \)[/tex] is given by the concentrations of the products divided by the concentrations of the reactants, each raised to the power of their respective coefficients in the balanced equation.
[tex]\[
K_c = \frac{[H_2O][CO]}{[H_2][CO_2]}
\][/tex]
Substitute the given equilibrium concentrations into this expression:
[tex]\[
[CO_2] = 9.5 \times 10^{-4} \, \text{M}
\][/tex]
[tex]\[
[H_2] = 4.5 \times 10^{-2} \, \text{M}
\][/tex]
[tex]\[
[H_2O] = 4.6 \times 10^{-3} \, \text{M}
\][/tex]
[tex]\[
[CO] = 4.6 \times 10^{-3} \, \text{M}
\][/tex]
So,
[tex]\[
K_c = \frac{(4.6 \times 10^{-3})(4.6 \times 10^{-3})}{(4.5 \times 10^{-2})(9.5 \times 10^{-4})}
\][/tex]
This calculation results in:
[tex]\[
K_c = \frac{(4.6 \times 10^{-3})^2}{(4.5 \times 10^{-2})(9.5 \times 10^{-4})}
\][/tex]
[tex]\[
K_c = \frac{2.116 \times 10^{-5}}{4.275 \times 10^{-5}}
\][/tex]
[tex]\[
K_c \approx 0.495
\][/tex]
Converting this to scientific notation, we obtain:
[tex]\[
K_c \approx 4.9 \times 10^{-1}
\][/tex]
Therefore, the value of the equilibrium constant for this reaction is [tex]\( 4.9 \times 10^{-1} \)[/tex]. So, the correct answer is:
[tex]\[
4.9 \times 10^{-1}
\][/tex]
