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

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Alleei Alleei · Answer 1 · 2019-08-07T12:57:45+02:00

Answer : The value of [tex]K'_c[/tex] for the reaction is, 7.5

Explanation :

Initial concentration of [tex]N_2O_4[/tex] = [tex]\frac{Moles}{Volume}=\frac{0.625}{5}=0.125M[/tex]

The balanced decomposition equilibrium reaction is,

[tex]N_2O_4(g)\rightleftharpoons 2NO_2(g)[/tex]

Initial conc. 0.125 0

At eqm. (0.125-x) 2x

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{(2x)^2}{(0.125-x)}[/tex]

The concentration of [tex]N_2O4[/tex] 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:

[tex]K_c=\frac{(2\times 0.05)^2}{(0.0750)}[/tex]

[tex]K_c=0.133[/tex]

Now we have to calculate the value of [tex]K_c[/tex] for the given reaction.

[tex]2NO_2(g)\rightleftharpoons N_2O_4(g)[/tex] [tex]K'_c[/tex]

As we know that,

[tex]K'_c=\frac{1}{K_c}[/tex]

So,

[tex]K'_c=\frac{1}{0.133}[/tex]

[tex]K'_c=7.5[/tex]

Therefore, the value of [tex]K'_c[/tex] for the reaction is, 7.5