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.
