Question

The brown gas NO2 and the colorless gas N2O4 exist in equilibrium,2NO2 N2O4.In an experiment, 0.625 mole of N2O4 was introduced into a 5.00 L vessel and was allowed to decompose until equilibrium was reached. The concentration of N2O4 at equilibrium was 0.0750 M. Calculate Kc for the reaction.0.0500.07500.100.1257.5

Answer 1

Answer : The value of $K'_c$ for the reaction is, 7.5

Explanation :

Initial concentration of $N_2O_4$ = $\frac{Moles}{Volume}=\frac{0.625}{5}=0.125M$

The balanced decomposition equilibrium reaction is,

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

Initial conc.     0.125          0

At eqm.          (0.125-x)       2x

The expression of equilibrium constant for the reaction will be:

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

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

$K_c=\frac{(2x)^2}{(0.125-x)}$

The concentration of $N_2O4$ at equilibrium = 0.0750 M

So, 0.125 - x = 0.0750

x = 0.05 M

Now put the value of 'x' in the above expression, we get:

$K_c=\frac{(2\times 0.05)^2}{(0.0750)}$

$K_c=0.133$

Now we have to calculate the value of $K_c$ for the given reaction.

$2NO_2(g)\rightleftharpoons N_2O_4(g)$   $K'_c$

As we know that,

$K'_c=\frac{1}{K_c}$

So,

$K'_c=\frac{1}{0.133}$

$K'_c=7.5$

Therefore, the value of $K'_c$ for the reaction is, 7.5