Question

Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and is allowed to reach equilibriumdescribed by the equation N2O4(g)↔ 2NO2(g). If at equilibrium theN2O4is 40.% dissociated, what is the value of the equilibriumconstant (in units of moles per liter) for the reaction under theseconditions?

a. 0.20
b. 0.84
c. 1.1
d. 1.5
e. 2.0

Answer 1

Answer : The correct option is, (C) 1.1

Solution : Given,

Initial moles of
`N_2O_4` = 1.0 mole

Initial volume of solution = 1.0 L

First we have to calculate the concentration
`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 given equilibrium reaction is,

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

Initially c 0

At equilibrium
`(c-c\alpha)`
`2c\alpha`

The expression of
`K_c` will be,

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

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

where,

`\alpha` = degree of dissociation = 40 % = 0.4

Now put all the given values in the above expression, we get:

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

`K_c=((2* 1* 0.4)^2)/((1-1* 0.4))`

`K_c=1.066\aprrox 1.1`

Therefore, the value of equilibrium constant for this reaction is, 1.1