Chemistry: Exactly 1.0 mol n2o4 is placed in an empty 1.0-l container and is allowed to reach equilibrium described by the equation n2o4(g) ↔ 2no2(g) if at equilibrium the n2o4 is 37% dissociated, what is the value of the equilibrium constant, kc, for the reaction under these conditions?

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Exactly 1.0 mol n2o4 is placed in an empty 1.0-l container

gstevens

30.12.2021

Answer : The equilibrium constant K_c for the reaction is, 0.869

Explanation :

First we have to calculate the concentration of N_2O_4 .

$\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}$

$\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M$

The balanced equilibrium reaction is,

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

Initial conc. C 0

At eqm. conc. $(C-C\alpha)$ $(2C\alpha)$

As we are given,

The percent of dissociation = $\alpha$ = 37 % = 0.37

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be :

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

$K_c=\frac{(2C\alpha)^2}{(C-C\alpha)}$

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

$K_c=\frac{(2\times 1.0\times 0.37)^2}{(1.0-1.0\times 0.37)}$

K_c=0.869

Therefore, the equilibrium constant K_c for the reaction is, 0.869

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