Answer :
Final answer:
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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