Answer:
Equilibrium concentration of [tex]Cl_2O[/tex] is 0.870 M.
Explanation:
Initial concentration of water = [tex][H_2O]=1.00 M[/tex]
Initial concentration of [tex]Cl_2O=[Cl_2O]=1.00 M[/tex]
[tex]H_2O(g) + Cl_2O(g)\rightarrow 2HOCl(g), Kc = 0.0900[/tex]
initially
1.00 M 1.00 M 0
At equilibrium
(1.00m-x) M (1.00-x)M 2x
The expression of equilibrium constant is given by :
[tex]K_c=\frac{[HOCl]^2}{[H_2O][Cl_2O]}[/tex]
[tex]0.0900=\frac{(2x)^2}{(1.00-x)(1.00-x)}[/tex]
Solving for x :
x = 0.130
Equilibrium concentration of [tex]Cl_2O[/tex]:
[tex][Cl_2O]=(1,00-x) M=1.00 M-0.130 M=0.870 M[/tex]
