Respuesta :
Answer : The equilibrium constant [tex]K_c[/tex] for the reaction is, 0.869
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]
The balanced equilibrium reaction is,
[tex]N_2O_4(g)\rightleftharpoons 2NO_2(g)[/tex]
Initial conc. C 0
At eqm. conc. [tex](C-C\alpha)[/tex] [tex](2C\alpha)[/tex]
As we are given,
The percent of dissociation = [tex]\alpha[/tex] = 37 % = 0.37
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]
[tex]K_c=\frac{(2C\alpha)^2}{(C-C\alpha)}[/tex]
Now put all the values in this expression, we get :
[tex]K_c=\frac{(2\times 1.0\times 0.37)^2}{(1.0-1.0\times 0.37)}[/tex]
[tex]K_c=0.869[/tex]
Therefore, the equilibrium constant [tex]K_c[/tex] for the reaction is, 0.869
