Answer: Concentration of  = 0.328 M
 = 0.328 M 
Concentration of  = 0.328 M
 = 0.328 M
Concentration of  = 0.532 M
 = 0.532 M
Explanation:
Moles of   and
 and  = 0.430 mole
 = 0.430 mole
Volume of solution = 0.500 L
Initial concentration of  and
 and  =
  =
The given balanced equilibrium reaction is,
                             
Initial conc.          0.860M     0.860M           0
At eqm. conc.    (0.860-x) M  (0.860-x) M     (x) M
The expression for equilibrium constant for this reaction will be,
![K_c=\frac{[COCl_2]}{[CO][Cl_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCOCl_2%5D%7D%7B%5BCO%5D%5BCl_2%5D%7D)
Now put all the given values in this expression, we get :

By solving the term 'x', we get :
x =  0.532 M
Thus, the concentrations of  at equilibrium are :
 at equilibrium are :
Concentration of  = (0.860-x) M =(0.860-0.532) M = 0.328 M
 = (0.860-x) M =(0.860-0.532) M = 0.328 M 
Concentration of  =  (0.860-x) M  =  (0.860-0.532) M = 0.328 M
 =  (0.860-x) M  =  (0.860-0.532) M = 0.328 M
Concentration of  = x M = 0.532 M
 = x M = 0.532 M