For the dissociation reaction
`N_(2)OhArr2NO_(2)(g),`
the equilibrium constant `K_(P)` is `0.120` atm at `298 K` and total pressure of system is `2` atm. Calculate the degree of dissociation of `N_(2)O_(4)`.

Explanation

For the reaction
`N_(2)O(g)hArr2NO_(2)(g)`
`{:("At equilibrium",,1-alpha,,2alpha),("moles",,,,):}`
Let `apha` be the degree of dissociation and P is the total pressure then
Total number of moles `=2alpha+1-alpha=1+alpha`
`:. P_(N_(2)O)= "moles"` of `N_(2)OxxP_(total)`
`=((1-alpha)/(1+alpha))P`
and `P_(NO_(2))=((2alpha)/(1+apha))P`
`:. K_(P)=((P_(NO_(2))))/(P_(N_(2)O))=([((2alpha)/(1+alpha))P]^(2))/(((1-alpha)/(1+alpha))P)`
`=(4alpha^(2)P^(2))/(1+alpha)xx(1+alpha)/((1-alpha)P)=(4alpha^(2)P)/(1-alpha^(2)) ...(1)`
Given `K_(P)=0.120 "atm", P=2 "atm"`
Subtituting all the values in equation (i), we get
`0.120=(4alpha^(2)(2))/((1-alpha^(2)))=(8alpha^(2))/((1-alpha^(2)))`
`rArr 0.120(1-alpha^(2))=8alpha^(2)`
`:.` Degree of dissociation, `alpha=(0.120/8.12)^(1//2)=0.121`