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

Answer 1

Answer: The value of equilibrium constant is 0.997

Explanation:

We are given:

Percent degree of dissociation = 39 %

Degree of dissociation, $\alpha$ = 0.39

Concentration of $N_2O_4$, c = $\frac{1mol}{1L}=1M$

The given chemical equation follows:

                     $N_2O_4\rightleftharpoons 2NO_2$

Initial:                c             -

At Eqllm:         $c-c\alpha$      $2c\alpha$

So, equilibrium concentration of $N_2O_4=c-c\alpha =[1-(1\times 0.39)]=0.61M$

Equilibrium concentration of $NO_2=2c\alpha =[2\times 1\times 0.39]=0.78M$

The expression of $K_{c}$ for above equation follows:

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

Putting values in above equation, we get:

$K_{c}=\frac{(0.78)^2}{0.61}\\\\K_{c}=0.997$

Hence, the value of equilibrium constant is 0.997