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 39% dissociated, what is the value of the equilibrium constant, Kc, for the reaction under these conditions

The value of equilibrium constant is 0.997

Explanation:

We are given:

Percent degree of dissociation = 39 %

Degree of dissociation, = 0.39

Concentration of , c =

The given chemical equation follows:

Initial:                c             -

At Eqllm:              

So, equilibrium concentration of

Equilibrium concentration of

The expression of for above equation follows:

Putting values in above equation, we get:

Hence, the value of equilibrium constant is 0.997

i would say b

explanation:

but i think i would need more information than that