Answer: Concentration of
= 0.328 M
Concentration of
= 0.328 M
Concentration of
= 0.532 M
Explanation:
Moles of
and
= 0.430 mole
Volume of solution = 0.500 L
Initial concentration of
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 :
Concentration of
= (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
Concentration of
= x M = 0.532 M