To determine the equilibrium constant ([tex]\(K_c\)[/tex]) for the given reaction:
[tex]\[
H_2O(g) + Cl_2O(g) \leftrightarrow 2 HClO(g)
\][/tex]
we need to use the equilibrium concentrations of the reactants and products. The equilibrium constant expression for this reaction is given by:
[tex]\[
K_c = \frac{{[HClO]^2}}{{[H_2O][Cl_2O]}}
\][/tex]
Given the equilibrium concentrations:
[tex]\[
[H_2O] = 0.077\, M
\][/tex]
[tex]\[
[Cl_2O] = 0.077\, M
\][/tex]
[tex]\[
[HClO] = 0.023\, M
\][/tex]
we plug these values into the equilibrium expression:
[tex]\[
K_c = \frac{{(0.023)^2}}{{(0.077) \times (0.077)}}
\][/tex]
First, calculate the numerator:
[tex]\[
(0.023)^2 = 0.000529
\][/tex]
Next, calculate the denominator:
[tex]\[
(0.077) \times (0.077) = 0.005929
\][/tex]
Now, divide the numerator by the denominator to find [tex]\(K_c\)[/tex]:
[tex]\[
K_c = \frac{{0.000529}}{{0.005929}} \approx 0.089
\][/tex]
Therefore, the equilibrium constant [tex]\(K_c\)[/tex] for the reaction at this temperature is [tex]\(0.089\)[/tex]. The correct answer from the given options is [tex]\(0.089\)[/tex].
