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The equilibrium constant K(p) for the fo...

The equilibrium constant `K_(p)` for the following reaction is `4.5`
`N_(2)O_(4)(g)hArr2NO_(2)(g)` What would be the average molar mass `("in"g//mol)` of an equilibrium mixture of `N_(2)O_(4)` and `NO_(2)` formed by the dissociation of pure `N_(2)O_(4)` at a total pressure of `2` atm ?

A

69

B

57.5

C

80.5

D

85.5

Text Solution

Verified by Experts

The correct Answer is:
B
`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`
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