InterviewSolution
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InterviewSolution
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`N_(2)O_(4)` is `25%` dissociated at `37^(@)C` and `1 atm`. Calculate (i) `K_(p)` and (ii) the percentage dissociation at `0.1` atm and `37^(@)C`. |
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Answer» `{:(,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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