Respuesta :
Answer : The correct option is, (a) 0.44
Explanation :
First we have to calculate the concentration of [tex]N_2O_4[/tex].
[tex]\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}[/tex]
[tex]\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M[/tex]
Now we have to calculate the dissociated concentration of [tex]N_2O_4[/tex].
The balanced equilibrium reaction is,
[tex]N_2O_4(g)\rightleftharpoons 2NO_2(aq)[/tex]
Initial conc. 1.0 M 0
At eqm. conc. (1.0-x) M (2x) M
As we are given,
The percent of dissociation of [tex]N_2O_4[/tex] = [tex]\alpha[/tex] = 28.0 %
So, the dissociate concentration of [tex]N_2O_4[/tex] = [tex]C\alpha=1.0M\times \frac{28.0}{100}=0.28M[/tex]
The value of x = [tex]C\alpha[/tex] = 0.28 M
Now we have to calculate the concentration of [tex]N_2O_4\text{ and }NO_2[/tex] at equilibrium.
Concentration of [tex]N_2O_4[/tex] = 1.0 - x = 1.0 - 0.28 = 0.72 M
Concentration of [tex]NO_2[/tex] = 2x = 2 × 0.28 = 0.56 M
Now we have to calculate the equilibrium constant for the reaction.
The expression of equilibrium constant for the reaction will be:
[tex]K_c=\frac{[NO_2]^2}{[N_2O_4]}[/tex]
Now put all the values in this expression, we get :
[tex]K_c=\frac{(0.56)^2}{0.72}=0.44[/tex]
Therefore, the equilibrium constant [tex]K_c[/tex] for the reaction is, 0.44
