The equilibrium constant `K_(P)` for the reaction `N_(2)O_(4)(g) hArr 2NO_(2)(g)` is `4.5 atm`. What would be the average molar mass (in `g//mol`) of an equilibrium mixture at a total pressure of `2 atm` of `N_(2)O_(4)` and `NO_(2)` formed by the dissociation of pure `N_(2)O_(4)` ?

Options

A. 69
B. 57.5
C. 80.5
D. 85.5

Answer

B

Explanation

`underset(P-Palpha)underset(P)(N_(2))O_(4)(g)hArrunderset(2Palpha)(2NO_(2)(g))` is 4.5
`P-Palpha+2Palpha=2`
`P+Palpha=2`
`4.5=(4P^(2)alpha^(2))/(P(1-alpha))`
`4.5=(4Palpha^(2))/(1-alpha)`
`4.5(1-alpha)=4Palpha^(2)`
`P=(4.5(1-alpha))/(4alpha^(2))`
`P(1+alpha)=2`
`(4.5(1-alpha)(1+alpha))/(4alpha^(2))=2`
`4.5(1-alpha^(2))=8alpha^(2)`
`4.5-4.5alpha^(2)=8alpha^(2)`
`4.5=12.5alpha^(2)`
`alpha=sqrt((4.5)/(12.5))`
`alpha=0.6`
`alpha=(M-EMM)/(EMM(n-1))`
`0.6EMM=92-EMM`
`1.6EMM=92`
`EMM=(92)/(1.6)`
`=57.5`