College

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

\[ C_2H_4(g) + H_2O(g) \rightleftharpoons C_2H_5OH(g) \]

Given:
- [tex]$K_c = 9.0 \times 10^3$[/tex]
- [tex]$[C_2H_4]_{eq} = 0.015 \, M$[/tex]
- [tex]$[C_2H_5OH]_{eq} = 1.69 \, M$[/tex]

Answer :

Answer:

The equilibrium concentration of water vapor is 0.0125 M.

Explanation:

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

The concentration of ethene at an equilibrium = [tex][C_2H_4]=0.015M[/tex]

The concentration of water vapor at an equilibrium = [tex][H_2O]=?[/tex]

The concentration of ethanol at an equilibrium = [tex][C_2H_5OH]=1.69 M[/tex]

the equilibrium constant of the reaction, [tex]K_c=9.0\times 10^{3}[/tex]

The expression of an equilibrium constant will be written as:

[tex]K_c=\frac{[C_2H_5OH]}{[C_2H_4][H_2O]}[/tex]

[tex]9.0\times 10^3=\frac{1.69 M}{0.015 M\times [H_2O]}[/tex]

[tex][H_2O]=\frac{1.69 M}{0.015 M\times 9\timers 10^3}=0.0125 M[/tex]

The equilibrium concentration of water vapor is 0.0125 M.