The equilibrium constant, kc, for the following reaction is 1.29×10-2 at 600 k. cocl2(g) co(g) + cl2(g) calculate the equilibrium concentrations of reactant and products when 0.323 moles of cocl2(g) are introduced into a 1.00 l vessel at 600 k.

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,

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,

Now put all the given values in this expression, we get :

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.

2b2t

explanation:

2b2t

1 is the answer to this problem