Exactly 1.0 mol n2o4 is placed in an empty 1.0-l container and is allowed to reach equilibrium described by the equation n2o4(g) ↔ 2no2(g) if at equilibrium the n2o4 is 37% dissociated, what is the value of the equilibrium constant, kc, for the reaction under these conditions?

Answer : The equilibrium constant for the reaction is, 0.869

Explanation :

First we have to calculate the concentration of .

The balanced equilibrium reaction is,

Initial conc.        C                 0

At eqm. conc.                  

As we are given,

The percent of dissociation = = 37 %  = 0.37

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be :

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

Therefore, the equilibrium constant for the reaction is, 0.869

1.43 x 10^-11 j the guy commented the awnser just making sure u see it .