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 39% dissociated, what is the value of the equilibrium constant, Kc, for the reaction under these conditions

Answer 1

Answer: The value of equilibrium constant is 0.997

Step-by-step explanation:

We are given:

Percent degree of dissociation = 39 %

Degree of dissociation,
`\alpha` = 0.39

Concentration of
`N_2O_4`, c =
`(1mol)/(1L)=1M`

The given chemical equation follows:

`N_2O_4\rightleftharpoons 2NO_2`

Initial: c -

At Eqllm:
`c-c\alpha`
`2c\alpha`

So, equilibrium concentration of
`N_2O_4=c-c\alpha =[1-(1* 0.39)]=0.61M`

Equilibrium concentration of
`NO_2=2c\alpha =[2* 1* 0.39]=0.78M`

The expression of
`K_(c)` for above equation follows:

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

Putting values in above equation, we get:

`K_(c)=((0.78)^2)/(0.61)\\\\K_(c)=0.997`

Hence, the value of equilibrium constant is 0.997