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N(2)O(4) is 25% dissociated at 37^(@)C a...

`N_(2)O_(4)` is `25%` dissociated at `37^(@)C` and `1 atm`. Calculate `K_(p)`

Text Solution

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`{:(,N_(2)O_(4),hArr,2NO_(2)),("Initial",1,,0),("At equilibrium",(1-a),,2a):}`
Total moles`=1+a`
`P_(N_(2)O_(4))=((1-a))/((1+a))p`
`P_(NO_(2))=(2a)/((1+a))p`
`K_(p)=([NO_(2)]^(2))/([N_(2)O_(4)])=([(2a)/((1+a))p]^(2))/([((1-a))/((1+a))p])=(4a^(2)p)/(1-a^(2))`
Putting `a=0.25`, we get
`K_(p)=0.267`
Calculating the percentage of dissociation at a pressure of `0.1` atm, we get
`0.267=(4a^(2)xx0.1)/(1-a^(2))`
or `a~~0.63`
Percentage dissociation `=63%`
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