Question

calculate K for N2O4 <--> 2NO at the equilibrium temperature of -40C, a 0.15M sample of N2O4 undergoes a decomposition of 0.456%

Answer 1

Answer: The value of equilibrium constant at -40°C is
`1.26* 10^(-5)`

Step-by-step explanation:

We are given:

Percent degree of dissociation = 0.456 %

Degree of dissociation,
`\alpha` = 0.00456

Concentration of
`N_2O_4`, c = 0.15 M

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 =[0.15-(0.15* 0.00456)]=0.1493M`

Equilibrium concentration of
`NO_2=2c\alpha =[2* 0.15* 0.00456]=0.00137M`

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

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

Putting values in above equation, we get:

`K_(eq)=((0.00137)^2)/(0.1493)\\\\K_(eq)=1.26* 10^(-5)`

Hence, the value of equilibrium constant at -40°C is
`1.26* 10^(-5)`