Answer : The equilibrium concentrations of
are, 0.2646, 0.0584 and 0.0584.
Explanation : Given,
Moles of
= 0.323 mole
Volume of solution = 1 L
Initial concentration of
= 0.323 M
Let the moles of
be, 'x'. So,
Concentration of
= x M
Concentration of
= x M
The given balanced equilibrium reaction is,
![COCl_2(g)\rightleftharpoons CO(g)+Cl_2(g)](https://tex.z-dn.net/?f=COCl_2%28g%29%5Crightleftharpoons%20CO%28g%29%2BCl_2%28g%29)
Initial conc. 0.323 M 0 0
At eqm. conc. (0.323-x) M (x M) (x M)
The expression for equilibrium constant for this reaction will be,
![K_c=\frac{[CO][Cl_2]}{[COCl_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCO%5D%5BCl_2%5D%7D%7B%5BCOCl_2%5D%7D)
Now put all the given values in this expression, we get :
![1.29\times 10^{-2}=\frac{(x)\times (x)}{(0.323-x)}](https://tex.z-dn.net/?f=1.29%5Ctimes%2010%5E%7B-2%7D%3D%5Cfrac%7B%28x%29%5Ctimes%20%28x%29%7D%7B%280.323-x%29%7D)
By solving the term 'x', we get :
x = 0.0584
Thus, the concentrations of
at equilibrium are :
Concentration of
= (0.323-x) M = (0.323-0.0584) M = 0.2646 M
Concentration of
= x M = 0.0584 M
Concentration of
= x M = 0.0584 M
Therefore, the equilibrium concentrations of
are, 0.2646, 0.0584 and 0.0584.