Question

Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and allowed to reach equilibrium described by the equation N2O4(g) 2NO2(g). If at equilibrium the N2O4 is 28.0% dissociated, what is the value of the equilibrium constant, Kc, for the reaction under these conditions? a. 0.44 b. 2.3 c. 0.11 d. 0.78 e. 0.31

Answer 1

Answer : The correct option is, (a) 0.44

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=(1.0moles)/(1.0L)=1.0M`

Now we have to calculate the dissociated concentration of
`N_2O_4`.

The balanced equilibrium reaction is,

`N_2O_4(g)\rightleftharpoons 2NO_2(aq)`

Initial conc. 1.0 M 0

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

As we are given,

The percent of dissociation of
`N_2O_4` =
`\alpha` = 28.0 %

So, the dissociate concentration of
`N_2O_4` =
`C\alpha=1.0M* (28.0)/(100)=0.28M`

The value of x =
`C\alpha` = 0.28 M

Now we have to calculate the concentration of
`N_2O_4\text{ and }NO_2` at equilibrium.

Concentration of
`N_2O_4` = 1.0 - x = 1.0 - 0.28 = 0.72 M

Concentration of
`NO_2` = 2x = 2 × 0.28 = 0.56 M

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

The expression of equilibrium constant for the reaction will be:

`K_c=([NO_2]^2)/([N_2O_4])`

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

`K_c=((0.56)^2)/(0.72)=0.44`

Therefore, the equilibrium constant
`K_c` for the reaction is, 0.44