Question

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?

Answer 1

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

`K_c=((2C\alpha)^2)/((C-C\alpha))`

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

`K_c=((2* 1.0* 0.37)^2)/((1.0-1.0* 0.37))`

`K_c=0.869`

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