Respuesta :
Answer : The concentrations of [tex]HOCl[/tex], [tex]H_2O[/tex] and [tex]Cl_2O[/tex] at equilibrium is, 0.215 M, 0.856 M and 0.606 M respectively.
Explanation :
The given chemical reaction is:
[tex]H_2O(g)+Cl_2O(g)\rightarrow 2HOCl(g)[/tex]
Initial conc. 0.590 0.340 0.747
At eqm. (0.590-x) (0.340-x) (0.747+2x)
As we are given:
[tex]K_c=0.0900[/tex]
The expression for equilibrium constant is:
[tex]K_c=\frac{[HOCl]^2}{[H_2O][Cl_2O]}[/tex]
Now put all the given values in this expression, we get:
[tex]0.0900=\frac{(0.747+2x)^2}{(0.590-x)\times (0.340-x)}[/tex]
x = -0.520 and x = -0.266
We are neglecting the value of x = -0.520 because equilibrium concentration can not be more than initial concentration.
Thus, the value of x will be, -0.266
The concentrations of [tex]HOCl[/tex] at equilibrium = (0.747+2x) = (0.747+2(-0.266)) = 0.215 M
The concentrations of [tex]H_2O[/tex] at equilibrium = (0.590-x) = (0.590+0.266) = 0.856 M
The concentrations of [tex]Cl_2O[/tex] at equilibrium = (0.340-x) = (0.340+0.266) = 0.606 M
