Question

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 1

Answer : The equilibrium constant $K_c$ for the reaction is, 0.869

Explanation :

First we have to calculate the concentration of $N_2O_4$.

$\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}$

$\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M$

The balanced equilibrium reaction is,

                          $N_2O_4(g)\rightleftharpoons 2NO_2(g)$

Initial conc.        C                 0

At eqm. conc.     $(C-C\alpha)$               $(2C\alpha)$

As we are given,

The percent of dissociation = $\alpha$ = 37 %  = 0.37

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

The expression of equilibrium constant for the reaction will be :

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

$K_c=\frac{(2C\alpha)^2}{(C-C\alpha)}$

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

$K_c=\frac{(2\times 1.0\times 0.37)^2}{(1.0-1.0\times 0.37)}$

$K_c=0.869$

Therefore, the equilibrium constant $K_c$ for the reaction is, 0.869