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

The equilibrium constant `K_(p)` for the reaction
`N_(2)O_(4)(g)hArr2NO_(2)(g)`
at `497^(@)C` is found to be `636 mm Hg`. If the pressure of the gas mixture is `182 mm`, calculate the presentage dissociation of `N_(2)O_(4)`. At what pressure will it be dissociated?

Text Solution

Verified by Experts

For this reaction, the expression of equilibrium constant is given by:
`K_(p)=(4alpha^(2)P_(T))/(1-alpha)`
`636=(4alpha^(2)xx182)/(1-alpha^(2)), [[:' K_(p)=636 (given)],[P_(T)=182 mm Hg]]`
`636-636alpha^(2)=728alpha^(2)`
`1364alpha^(2)=636`
`alpha^(2)=636/1364=0.4663 rArr alpha=sqrt(0.4663)=0.6829`
`%` dissociation of `N_(2)O_(4)=0.6829xx100=68.29`
When the gas is half dissociated, `alpha=0.5`
Let the pressure be `P_(T)^(') mm Hg`
`:. 636=(4xx(0.5)^(2)xxP_(T)^('))/(1-(0.5)^(2))`
`P_(T)^(')=477 mm`
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