Question

one mole of N2O4 in a 1 lt flask decomposes to attain the equilibrium .At equilibrium the mole fraction of NO2 is 1/2 .Hence Kc will be​

Answer 1

Answer:

Explanation:i have just added the picture of solution

Answer 2

Answer: 0.67

Explanation:

Initial moles of  $N_2O_4$ = 1 mole

Mole fraction of  $NO_2$ at equilibrium= $\frac{1}{2}$

The balanced equilibrium reaction is,

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

Initial conc.            1            0

At eqm. conc.    (1-x)          2x

The expression for equilibrium constant for this reaction will be,

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

$K_c=\frac{(2x)^2}{1-x}$

Total moles at equilibrium = (1-x) + 2x = 1 + x

we are given :

mole fraction of $NO_2=\frac{\text {moles of }NO_2}{\text {total moles}}=\frac{2x}{1+x}$

$\frac{2x}{1+x}=\frac{1}{2}$

$x=\frac{1}{3}$

Equilibrium concentration of $N_2O_4=\frac{moles}{volume}=\frac{1-\frac{1}{3}}{1L}=0.67M$

Equilibrium concentration of $NO_2=\frac{moles}{volume}=\frac{\frac{2}{3}}{1L}=0.67M$

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

$K_c=\frac{(0.67)^2}{0.67}$

$K_c=0.67$

Thus the value of the equilibrium constant is 0.67