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

Reaction: [tex]C_2H_4(g) + H_2O(g) \rightleftharpoons C_2H_5OH(g)[/tex]

Equilibrium constant: [tex]K_c = 9.0 \times 10^3[/tex]

Equilibrium concentrations:
[C₂H₄]ₑq = 0.015 M
[C₂H₅OH]ₑq = 1.69 M

Calculate the equilibrium concentration of [tex]H_2O(g)[/tex].

The equilibrium concentration of H2O(g) in the reaction C2H4(g) + H2O(g) ⇌ C2H5OH(g) at the given equilibrium constant (Kc) of 9.0 × 10³ is 0.021 M, when calculated with the provided equilibrium concentrations of reactants and products.

Explanation:

To determine the equilibrium concentration of H₂O(g) for the reaction C₂H₄(g) + H₂O(g) ⇌ C₂H₅OH(g) with a given Kc of 9.0 × 10³, equilibrium concentrations of C₃H₄ and C₂H₅OH, and assuming that the C₃H₄ concentration is a typo for C₂H₄, we can set up an equilibrium expression.

Given the equilibrium constant expression Kc = [C₂H₅OH] / ([C₂H₄][H₂O]), and the provided equilibrium concentrations [C₂H₄]eq = 0.015 M and [C₂H₅OH]eq = 1.69 M, we can solve for the missing concentration of H₂O:

Kc × ([C₂H₄][H₂O]) = [C₂H₅OH]

(9.0 × 10³) × (0.015 M × [H₂O]) = 1.69 M

Solving for [H₂O], we find:

[H₂O] = 1.69 M / (9.0 × 10³ × 0.015 M)

[H₂O]eq = 0.021 M

Thus, the equilibrium concentration of H₂O(g) is 0.021 M.

This agrees with the provided equilibrium constant value when rounded to the appropriate number of significant figures, confirming our calculated concentrations.