Question

Plz help, I will give brainliest.

Answer 1

Answer: The value of $K_c$ is 2

Explanation:

Moles of  $N_2O_4$ = 1.0 mole  

Volume of solution = 1.00 L

Initial concentration of $N_2O_4$ = $\frac{1.0mol}{1.00L}=1.0M$

Equilibrium concentration of $NO_2$ = $\frac{1.0mol}{1.00L}=1.0M$  

The given balanced equilibrium reaction is,

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

Initial conc.          1.0 M            0 M

At eqm. conc.     (1.0-x) M      (2x) M

The expression for equilibrium constant for this reaction will be,

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

Given : 2x = 1.0

x= 0.5

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

$K_c=\frac{(1.0)^2}{(1.0-0.5)}$

$K_c=2$

Thus the value of $K_c$ is 2