Consider the following reaction, equilibrium concentrations, and equilibrium constant at a particular temperature. Determine the equilibrium concentration of [tex]H_2O(g)[/tex].

\[ \text{C}_2\text{H}_4(g) + H_2O(g) \leftrightarrow \text{C}_2\text{H}_5\text{OH}(g) \]
\[ K_c = 9.0 \times 10^3 \]
\[ [\text{C}_2\text{H}_4]_{eq} = 0.015 \, M \]
\[ [\text{C}_2\text{H}_5\text{OH}]_{eq} = 1.69 \, M \]

A. [tex]9.9 \times 10^{-7} \, M[/tex]
B. [tex]1.68 \, M[/tex]
C. [tex]80 \, M[/tex]
D. [tex]0.013 \, M[/tex]
E. [tex]1.0 \, M[/tex]

The equilibrium concentration of water in the given reaction under specified conditions, calculated using the law of mass action and given equilibrium concentrations, is 1.25 x 10^-5 M.

The equilibrium concentration of H2O(g) is 1.68M

Explanation:

The song is often used to solve problems involving

equilibrium concentrations

. According to the law of mass action, the equilibrium constant Kc for this reaction is given by Kc = [C2H5OH]/([C2H4][H2O]). Here, [C2H5OH] is the equilibrium concentration of C2H5OH and similarly for [C2H4] and [H2O]. We can substitute the given values into this equation to find the unknown, that is [H2O].

Therefore, [H2O] = [C2H5OH]/([C2H4]Kc) = 1.69M/(0.015M * 9.0×10^3) = 1.25 x 10^-5 M.

Let's be professional and friendly, this tells us that the equilibrium concentration of water in this reaction, under these conditions, is 1.25 x 10^-5 M.

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